2. In a quadrilateral field ABCD, ZB = 90°, AB = 80 m, BC = 150 m, CD

1.	2. In a quadrilateral field ABCD, ZB = 90°, AB = 80 m, BC = 150 m, CD = 120 m andDA = 250 m. Find the area of the field in hectares. Find the price of the field ifit cost12,000 / hectare.
Answer» $\huge \bold \color{green} \mid \mid{ \underline{ \underline \red{required \: answer}}} \mid \mid$ $\bold { {\underline{ \underline \orange{given}}}}$ angle B = 90° AB = 80m BC = 150m CD = 120m DA = 250m $\bold { {\underline{ \underline \orange{answer}}}} \:$ Rs. 18,000 $\bold { {\underline{ \underline \orange{solution}}}} \:$ In RIGHT angle ∆ ABC AB² + BC² = AC² ( by Pythagoras theorem ) => (80)² + (150)² = AC² => 6400 + 22500 = AC² => 28900 = AC => AC = 170m $ar( \triangle) = \frac{<klux>1</klux>}{2} \times b \times h \\ \\ \implies ar(ABC) = \frac{1}{2} \times 150 \times 80 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 150 \times 40 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6000 {m}^{2} ............ar(i)$ now we have to find area of ∆ACD by heron's formula $ar( \triangle) = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ s = \frac{a + b + c}{2}$ so in triangle ACD $s = \frac{AC + CD + DA}{2} \\ \\ s = \frac{170 + 120 + 250}{2} \\ \\ s = \frac{540}{2} \\ \\ s = 270$ then area of triangle ACD, $ar(acd) = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{270(270 - 250)(270 - 120)(270 - 170)} \\ \\ = \sqrt{270 \times 20 \times 150 \times 100} \\ \\ = \sqrt{27 \times 10 \times 2 \times 10 \times 15 \times 10 \times 10 \times 10} \\ \\ = 100 \sqrt{27 \times 2 \times15 \times 10 } \\ \\ = 100 \sqrt{3 \times 3 \times 3 \times 2 \times 3 \times 5 \times 2 \times 5} \\ \\ = 100 \times 3 \times 3 \times 2 \times 5 \\ \\ = 9000 {m}^{2} ............ar(ii)$ now area of quadrilateral ABCD is; ar(ABCD) = ar(ABC) + ar(ACD) $\implies \: ar(abcd) = 6000 + 9000 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 15000 {m}^{2}$ so area of quadrilateral field is 15000m² we know that.. 1 hectare = SQUARE metre / 10000 $\bold {\:so \: 15000 {m}^{2} = 1.5hectare}$ so area of field is 1.5 Hectare cost = 12,000 / hectare then price of field is; $\bold{1.5 \times cost \: of \: field} \\ \\ \bold{\implies \: price \: = 1.5 \times 12000} \\ \\ \bold{\implies \: price \: of \: field \: = <klux>18000</klux> \: rs }$ hence, price of the field is Rs. 18000

2. In a quadrilateral field ABCD, ZB = 90°, AB = 80 m, BC = 150 m, CD = 120 m andDA = 250 m. Find the area of the field in hectares. Find the price of the field ifit cost12,000 / hectare.

Answer»

$\huge \bold \color{green} \mid \mid{ \underline{ \underline \red{required \: answer}}} \mid \mid$

$\bold { {\underline{ \underline \orange{given}}}}$

angle B = 90°

AB = 80m

BC = 150m

CD = 120m

DA = 250m

$\bold { {\underline{ \underline \orange{answer}}}} \:$

Rs. 18,000

$\bold { {\underline{ \underline \orange{solution}}}} \:$

In RIGHT angle ∆ ABC

AB² + BC² = AC² ( by Pythagoras theorem )

=> (80)² + (150)² = AC²

=> 6400 + 22500 = AC²

=> 28900 = AC

=> AC = 170m

$ar( \triangle) = \frac{<klux>1</klux>}{2} \times b \times h \\ \\ \implies ar(ABC) = \frac{1}{2} \times 150 \times 80 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 150 \times 40 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6000 {m}^{2} ............ar(i)$

now we have to find area of ∆ACD by heron's formula

$ar( \triangle) = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ s = \frac{a + b + c}{2}$

so in triangle ACD

$s = \frac{AC + CD + DA}{2} \\ \\ s = \frac{170 + 120 + 250}{2} \\ \\ s = \frac{540}{2} \\ \\ s = 270$

then area of triangle ACD,

$ar(acd) = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{270(270 - 250)(270 - 120)(270 - 170)} \\ \\ = \sqrt{270 \times 20 \times 150 \times 100} \\ \\ = \sqrt{27 \times 10 \times 2 \times 10 \times 15 \times 10 \times 10 \times 10} \\ \\ = 100 \sqrt{27 \times 2 \times15 \times 10 } \\ \\ = 100 \sqrt{3 \times 3 \times 3 \times 2 \times 3 \times 5 \times 2 \times 5} \\ \\ = 100 \times 3 \times 3 \times 2 \times 5 \\ \\ = 9000 {m}^{2} ............ar(ii)$

now area of quadrilateral ABCD is;

ar(ABCD) = ar(ABC) + ar(ACD)

$\implies \: ar(abcd) = 6000 + 9000 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 15000 {m}^{2}$

so area of quadrilateral field is 15000m²

we know that..

1 hectare = SQUARE metre / 10000

$\bold {\:so \: 15000 {m}^{2} = 1.5hectare}$

so area of field is 1.5 Hectare

cost = 12,000 / hectare

then price of field is;

$\bold{1.5 \times cost \: of \: field} \\ \\ \bold{\implies \: price \: = 1.5 \times 12000} \\ \\ \bold{\implies \: price \: of \: field \: = <klux>18000</klux> \: rs }$

hence, price of the field is Rs. 18000

2. In a quadrilateral field ABCD, ZB = 90°, AB = 80 m, BC = 150 m, CD = 120 m andDA = 250 m. Find the area of the field in hectares. Find the price of the field ifit cost12,000 / hectare.

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2. In a quadrilateral field ABCD, ZB = 90°, AB = 80 m, BC = 150 m, CD = 120 m andDA = 250 m. Find the area of the field in hectares. Find the price of the field ifit cost12,000 / hectare.​

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2. In a quadrilateral field ABCD, ZB = 90°, AB = 80 m, BC = 150 m, CD = 120 m andDA = 250 m. Find the area of the field in hectares. Find the price of the field ifit cost12,000 / hectare.