This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 601. |
Two parallel lines meet each other at some point. |
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Answer» The correct answer is False. |
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| 602. |
Fill in the blanks to make the statements true:In Fig., points A, B, C, D and E are collinear such that AB = BC = CD = DE. Then(a) AD = AB + ______(b) AD = AC + ______(c) mid point of AE is ______(d) mid point of CE is ______(e) AE = ______ × AB. |
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Answer» (a) AD = AB + BD Where, BD = BC + CD (b) AD = AC + CD (c) mid point of AE is c (d) mid point of CE is D (e) AE = 4 × AB. [given] |
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| 603. |
Fill in the blank,In Fig. 9.1, the line segments PQ and RQ have been marked on a line l such that PQ = AB and RQ = CD. Then AB – CD =__________. |
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Answer» Then AB – CD = PR |
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| 604. |
In the following figure, PQ is parallel to BC and `PQ:BC =1 :3`. If the area of the triangle ABC is `144 cm^2`, then what is the area of the triangle APQ? A. `48 cm ^2`B. `36cm ^2`C. `16 cm ^2`D. `9cm^2` |
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Answer» Correct Answer - C The ratio of areas of two similar triangles equal to the ratio of the squares of their corresponding sides. |
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| 605. |
A cyclic polygon has n sides such that each of its interior angle measures `144^(@)`. What is the measures of the angle subtended by each of its side at the geometrical center of the polygon.A. `144^(@)`B. `30^(@)`C. `36^(@)`D. `54^(@)` |
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Answer» Correct Answer - C Find the number of sides of the polygon, anlge subtended at the center `=360^(@)/n`. |
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| 606. |
In `DeltaPQR and Delta ABC`, `angleQ=angleB=90^(@),PQ=AB, and QR=BC`. Which of the following property can be used to prove the congruence of `DeltaPQR` and `DeltaABC` ?A. SSSB. RHSC. ASAD. SAS |
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Answer» Correct Answer - D Given `PQ=AB, QR=BC, angle Q=angleB=90^(@)`. Basically by using SAS congruence property. ` DeltaPQR~=ABC`. Hence, the correct option is (d). |
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| 607. |
In two triangles ABC an DEF,`angle A=angleD`. The sum of the angles A and B is equal to the sum of the angles D and E. If `BC=6` cm and `EF=8cm`, find the ratio of the areas of the triangles, ABC and DEF.A. `3:4 `B. `4:3`C. `9:16`D. `16:9` |
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Answer» Correct Answer - C The ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. |
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| 608. |
A man travels on a bicycle, 10 km east from the starting point A to reach point B, then the cycles 15 km south to reach point C. Find the shortest distance between A and C.A. 25 kmB. 5kmC. `25sqrt(13) km`D. `5sqrt(13) km` |
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Answer» Correct Answer - D Apply Pythagoras theorem. |
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| 609. |
In the following figure, PQ is parallel to BC and `PQ:BC =1 :3`. If the area of the triangle ABC is `144 cm^2`, then what is the area of the triangle APQ? A. `48 cm ^2`B. `36cm ^2`C. `16 cm ^2`D. `9cm^2` |
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Answer» Correct Answer - C The ratio of areas of two similar triangles equal to the ratio of the squares of their corresponding sides. |
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| 610. |
In triangle ABC , sides AB and AC are extended to D and E, respectively, such that `AB=BD` and AC =CE`, Find DE, if `BC=6 cm`.A. 3 cmB. 6cmC. 9 cmD. 12 cm |
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Answer» Correct Answer - C Apply basic proportionality theorem. |
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| 611. |
Using the information given, name the right angles in each part of Fig.is RS⊥RW? |
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Answer» The correct answer is ∠RTW and ∠RTS |
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| 612. |
Using the information given, name the right angles in each part of Fig.AC⊥ CD? |
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Answer» The correct answer is ∠ACD. |
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| 613. |
Using the information given, name the right angles in each part of Fig.AE⊥ CE? |
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Answer» The correct answer is ∠AEC. |
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| 614. |
In Fig., ∠BAC = 90° and AD ⊥ BC. The number of right triangles in the figure is(A) 1 (B) 2 (C) 3 (D) 4 |
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Answer» (C) 3 We have, ∠BAC = 90° and AD ⊥ BC ∵ ∠BDA = ∠CDA = ∠BAC = 90° ∴ There are 3 right triangles formed in the given figure. |
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| 615. |
A polygon has prime number of sides. Its number of sides is equal to the sum of the two least consecutive primes. The number of diagonals of the polygon is(A) 4 (B) 5 (C) 7 (D) 10 |
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Answer» (B) 5 The two least consecutive prime numbers are, 2 and 3 Sum of two numbers = 2 + 3 = 5 By using formula = n(n – 3)/2 = 5(5 – 3)/2 = (5 × 2)/2 = 10/2 = 5 |
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| 616. |
Using the information given, name the right angles in each part of Fig.AC⊥ BD? |
| Answer» The correct answer is ∠AED,∠AEB,∠BEC and ∠DEC. | |
| 617. |
Using the information given, name the right angles in each part of Fig.is AC⊥ BD? |
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Answer» The correct answer is ∠ACD and ∠ACB |
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| 618. |
In Fig., AB = BC and AD = BD = DC. The number of isosceles triangles in the figure is(A) 1 (B) 2 (C) 3 (D) 4 |
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Answer» (C) 3 ∵ We have, AB = BC and AD = BD = DC. ∴ ∆ ABD, ∆ BDC and ∆ ABC all are isosceles triangles. ∴ There are 3 isosceles triangles formed in the given figure. |
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| 619. |
In Fig, if AC ⊥ BD , then name all the right angles. |
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Answer» There are four right angles. They are: ∠APD , ∠APB, ∠BPC and ∠CPD. |
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| 620. |
Find out the incorrect statement, if any, in the following: An angle is formed when we have (a) two rays with a common end-point (b) two line segments with a common end-point (c) a ray and a line segment with a common end-point |
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Answer» The correct answer is (b) and (c) |
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| 621. |
In Fig, PQ ⊥ AB and PO = OQ. Is PQ the perpendicular bisector of line segment AB? Why or why not? |
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Answer» PQ is not the perpendicular bisector of line segment AB, because AO ≠ BO. [Note: AB is the perpendicular bisector of line segment PQ]. |
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| 622. |
In Fig. 2.1, PQ ⊥ AB and PO = OQ. Is PQ the perpendicular bisector of line segment AB? Why or why not? |
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Answer» PQ is not the perpendicular bisector of line segment AB, because AO ≠ BO. [Note: AB is the perpendicular bisector of line segment PQ]. PQ is not the perpendicularbisector of line segment AB, because AO ≠ BO. [Note: AB is the perpendicular bisector of line segment PQ]. |
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| 623. |
If line PQ || line m, then line segment PQ || m |
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Answer» The correct answer is True. |
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| 624. |
The number of independent measurement required to construct an isosceles trapezium is three. |
| Answer» Correct Answer - True | |
| 625. |
Number of independent measurement required to construct a triangle is ______ |
| Answer» Correct Answer - 3 | |
| 626. |
Construct a parallelogram PQRS, when `PQ=3.7 cm, QR=2.3cm, and PR=4.8cm`. |
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Answer» Steps: 1. Draw a line segment PQ=3.7 cm. 2. Draw an arc with P as the centre and a radius of 4.8 cm. 3. With Q as the centre and a radius of 2.3 cm, draw another arc to intersect the previous are of step 2 at R and join QR. 4. With R as the centre , draw an arc of radius 3.7 cm. 5. With P as the centre, draw another arc of radius 2.3 cm to intersect the arc in step 4 at S. Join RS and PS. PQRS is the required parallelogram. |
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| 627. |
In Fig. 2.39, (a) What is AE + EC? (b) What is AC – EC? (c) What is BD – BE? (d) What is BD – DE? |
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Answer» (a) AC (b) AE (c) ED (d) BE |
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| 628. |
In the adjoining figure, O is the centre of the circle. AB is an arc of the circle and `/_AOB=80^(@)`. Find `/_ACB`. |
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Answer» The angle made by an arc at the centre of a circle is twice the angle made by the arc point on the remaining part of the circle. `/_AOB=2 /_ACB` `implies 2 /_ACB=80^(@)` `implies /_ACB =40^(@)` |
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| 629. |
The number of independent measurement required to construct a circle is two. |
| Answer» Correct Answer - False | |
| 630. |
Centroid of a triangle divides its median in the ratio of 1:2 from the vertex. |
| Answer» Correct Answer - False | |
| 631. |
Find out the incorrect statement, if any, in the following: An angle is formed when we have (a) two rays with a common end-point (b) two line segments with a common end-point (c) a ray and a line segment with a common end-point |
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Answer» All the three statements (a), (b) and (c) are incorrect. ∵ The common initial point of two rays forms an angle. |
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| 632. |
The point of concurrence of altitudes of a triangle is called orthocentre. |
| Answer» Correct Answer - True | |
| 633. |
In which of the following figures,(a) perpendicular bisector is shown?(b) bisector is shown?(c) only bisector is shown?(d) only perpendicular is shown? |
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Answer» (a) Figure (ii) shows the perpendicular bisector (b) Figure (ii) and (iii) shows the bisector. (c) Figure (iii) shows only the bisector. (d) Figure (i) shows only the perpendicular. |
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| 634. |
The point of concurrence of medians of a triangle is called centroid. |
| Answer» Correct Answer - True | |
| 635. |
The image of a point lying on a line l with respect to the line of symmetry l lies on _______. |
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Answer» The image of a point lying on a line l with respect to the line of symmetry l lies on ɭ |
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| 636. |
Given below are some figures. Choose the image of the given figure with respect to the given line from the given choices. A. B. C. D. |
| Answer» Correct Answer - a | |
| 637. |
What is common in the following figures (i) and (ii)?Is figure (i) that of triangle ? if not, why? |
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Answer» Both the figures (i) and (ii) have 3 line segments. |
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| 638. |
If two rays intersect, will their point of intersection be the vertex of an angle of which the rays are the two sides? |
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Answer» Yes If two rays intersect, will their point of intersection be the vertex of an angle of which the rays are the two sides. |
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| 639. |
In given figure,(a) is AC + CB = AB ?(b) is AB + AC= CB ?(c) is AB + BC = CA ? |
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Answer» (a) Yes, AC + CB = AB (b) No, AB – AC = CB (c) No, AB – BC = CA |
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| 640. |
In Fig. 2.38, (a) is AC + CB = AB? (b) is AB + AC = CB? (c) is AB + BC = CA? |
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Answer» (a) Yes (b) No (c) No . |
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| 641. |
In given figure,(a) name any four angles that appear to be acute angles.(b) name any two angles that appear to be obtuse angles. |
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Answer» (a) Acute angles : ∠ADE, ∠AEB, ∠ABE and ∠ECD. (b) Obtuse angles : ∠BCD and ∠BAD. |
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| 642. |
In the following, CDEF is a cyclic quadrilateral. `bar(CG)` and `bar(DH)` are the angle bisectors of `angleC` and `angleD` respectively. If `angleE=100^(@)` and `angleF=110^(@)`, then find `angleCPD`. |
| Answer» In a cyclic quadrilateral, opposite angles are supplementary. | |
| 643. |
The angle subtended by a minor arc in its alternate segment is…………….A. acuteB. obtuseC. `90^(@)`D. reflex angle. |
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Answer» Correct Answer - A Angle subtended by a minor arc is always less than `90^(@)`. |
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| 644. |
A tower of height 60 m casts a 40 m long shadow on the ground. At the same time, a needle of height 12 m casts a `x` cm long shadow the ground. Find `x`A. 6B. 8C. 10D. 14 |
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Answer» Correct Answer - B Join,BD, ABDE is a cyclic quadrilateral. |
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| 645. |
A pole of height 14 m casts a 10 m long shadow on the ground. At the same time, a tower casts a 70 m long shadow on the ground. Find the height of the tower.A. 50 mB. 78 mC. 90 mD. 98 m |
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Answer» Correct Answer - D Use the concept of similar triangles. |
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| 646. |
Find the distance between the helicopter and the ship. |
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Answer» From the figure AS is the distance between the helicopter and the ship. ∆ APS is a right angled triangle, by Pythagoras theorem, AS2 = AP2 + PS2 = 802 + 1502 = 6400 + 22500 = 28900 = 1702 ∴ The distance between the helicopter and the ship is 170 m |
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| 647. |
How many edges, faces and vertices are there in a sphere? |
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Answer» No edges, No faces and No vertices. |
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| 648. |
ABCD is a rhombus, in which the length of the diagonals AC and BD are 6 cm and 8cm, respectively. Find the perimeter of the rhombus ABCD. |
| Answer» Correct Answer - 20 cm | |
| 649. |
In a parallelogram PQRS, the bisectors of `/_P` and `/_Q` meet on RS.If the perimeter PQRS is 13.5 cm, then find the measure of QRA. 4.5 cmB. 2.25 cmC. 3 cmD. 3.75cm |
| Answer» Correct Answer - b | |
| 650. |
If the three angles of a triangle are in the ratio 3 : 5 : 4, then find them. |
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Answer» Given three angles of the triangles are in the ratio 3 : 5 : 4. Let the three angle be 3x, 5x and 4x. By angle sum property of a triangle, we have 3x + 5x + 4x = 180° 12x = 180° x = \(\frac{180°}{12}\) x = 15° ∴ The angle are 3x = 3 x 15° = 45° 5x = 5 x 15° = 75° 4x = 4 x 15° = 60° Three angles of the triangle are 45°, 75°, 60° |
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