Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN

1.	Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.
Answer» Given: A pentagon ABCDE BL \(\perp\) AC, CM \(\perp\) AD and EN \(\perp\) AD AC = 10 cm AD = 12 cm BL = 3 cm CM = 7 cm EN = 5 cm Here, Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE) Area of triangle = \(\frac{1}{2}\)× (base) × (height). Here, Area (\(\triangle\)ABC) = \(\frac{1}{2}\)× (AC) × (BL) = \(\frac{1}{2}\)× (10) × (3) = 15 cm². Area (\(\triangle\)ACD) = \(\frac{1}{2}\)× (AD) × (CD) = \(\frac{1}{2}\)× (12) × (7) = 42 cm². Area (\(\triangle\)ADE) = \(\frac{1}{2}\)× (AD) × (EN) = \(\frac{1}{2}\)× (12) × (5) = 30 cm². ∴ Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE) = 15 + 42 + 30 = 87 cm². ∴ Area (Pent. ABCDE) = 87 cm².

Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.

Answer»

Given: A pentagon ABCDE

BL \(\perp\) AC, CM \(\perp\) AD and EN \(\perp\) AD

AC = 10 cm

AD = 12 cm

BL = 3 cm

CM = 7 cm

EN = 5 cm

Here,

Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE)

Area of triangle = \(\frac{1}{2}\)× (base) × (height).

Here,

Area (\(\triangle\)ABC) = \(\frac{1}{2}\)× (AC) × (BL) = \(\frac{1}{2}\)× (10) × (3) = 15 cm².

Area (\(\triangle\)ACD) = \(\frac{1}{2}\)× (AD) × (CD) = \(\frac{1}{2}\)× (12) × (7) = 42 cm².

Area (\(\triangle\)ADE) = \(\frac{1}{2}\)× (AD) × (EN) = \(\frac{1}{2}\)× (12) × (5) = 30 cm².

∴ Area (Pent. ABCDE) = area (\(\triangle\)ABC) + area (\(\triangle\)ACD) + area (\(\triangle\)ADE) = 15 + 42 + 30 = 87 cm².

∴ Area (Pent. ABCDE) = 87 cm².