Find the area of a parallelogram given in Fig. 12.2. Also find the length of the altitude from vertex A on the side DC.(in $c m^{2}$ )

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Solution

$W e k n o w t h a t t h e d i a g o n a l o f a p a r a l l e l o g r a m (∥^{g m}) d i v i d e s i t i n t o t w o c o n g r u e n t t r i a n g l e s . S o A r e a o f ∥^{g m} A B C D = 2 \times A r e a o f △ B C D . A c c o r d i n g t o H e r o n^{'} s f o r m u l a t h e a r e a (A) o f t r i a n g l e w i t h s i d e s a, b & c i s g i v e n a s A = \sqrt[2]{[s (s - a) (s - b) (s - c)]} w h e r e 2 s = (a + b + c) . H e r e a = 12, b = 17, c = 25, s = \frac{(12 + 17 + 25)}{2} = \frac{54}{2} = 27 A r e a o f ∥^{g m} A B C D = 2 \times \sqrt[2]{[27 \times (27 - 12) (27 - 17) (27 - 25)]} = 2 \times \sqrt[2]{(27 \times 15 \times 10 \times 2)} = 2 \times \sqrt[2]{(3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2)} = 2 \times \sqrt[2]{[(3 \times 3 \times 5 \times 2) (3 \times 3 \times 5 \times 2)]} = 2 \times (3 \times 3 \times 5 \times 2) = 2 \times 90 = 180 A r e a o f ∥^{g m} = b a s e \times h e i g h t H e i g h t o f a l t i t u d e f r o m v e r t e x A o n s i d e C D o f t h e o f ∥^{g m} = \frac{a r e a o f ∥^{g m} A B C D}{b a s e C D} = \frac{180}{12} = 15 c m$

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Question 5
Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC.

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12

A D = 8 c m

A B C D = 44

c m^{2}

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