In the given figure BEDC is a parallelogram If the area of - ADC - 8- area of - BDC - 3 and - DAB-90- then the area of - AEB is

The correct option is A 5 Area of a triangle = 12 height × base Area ADC = 12 AD × AB = 8 or AD × AB = 16 (1)Now, Area BDC = 12 ...

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Byju's Answer

Standard IX

Mathematics

Theorem 1: Diagonal of a Parallelogram Divides It into Two Congruent Triangles

In the given ...

Question

In the given figure BEDC is a parallelogram If the area of $Δ A D C$ = 8, area of $Δ B D C$ = 3 and $∠ D A B = 90^{\circ}$ , then the area of $Δ A E B$ is

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Solution

The correct option is A $5$
Area of a triangle = $\frac{1}{2} h e i g h t \times b a s e$
Area $△ A D C$ = $\frac{1}{2} A D \times A B = 8$
or $A D \times A B = 16$ (1)
Now, Area $△ B D C = \frac{1}{2} A B \times B C = 3$
$A B \times B C = 6$ (2)
Subtracting (1) and (2)
$A B (A D - B C) = 10$
$A B \times A E = 10$
$\frac{1}{2} A B \times A E = 5$
Hence, Ar. $△ A E B = 5$

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In the given figure BEDC is a parallelogram If the area of ΔADC = 8, area of ΔBDC = 3 and ∠DAB=90∘, then the area of ΔAEB is

The correct option is A 5Area of a triangle = 12height×baseArea △ADC = 12AD×AB=8or AD×AB=16 (1)Now, Area △BDC=12AB×BC=3AB×BC=6 (2)Subtracting (1) and (2)AB(AD−BC)=10AB×AE=1012AB×AE=5Hence, Ar. △AEB=5

In the given figure BEDC is a parallelogram If the area of $Δ A D C$ = 8, area of $Δ B D C$ = 3 and $∠ D A B = 90^{\circ}$ , then the area of $Δ A E B$ is