find the area of the figure
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Answer:
area of ∆BCD=1/2×b×h
1/2×16×10=80cm
area of ∆BAD=1/2×b×h
1/2×16×6=48cm
area of ABCD=80+48=128cm
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Answer:
Tha answer is as follows:
Step-by-step explanation:
The area of the figure = Area of ∆ABD + Area of ∆CBD
Base of both triangles is common i.e. 16 cm
Height of ∆ABD (h1) = 6 cm
Height of ∆CBD (h2) = 10 cm
Hence, Area of ∆ABD = 1 × (b.h) = 1 × 16 × 6 = 48 cm² ---- 1.
2 2
Area of ∆CBD = 1 × (b.h) = 1 × 16 × 10 = 80 cm² ---- 2.
2 2
A.T.Q,
Area of the figure> 1 × base × h1 + 1 × base × h2
2 2 [From 1. and 2.]
Area of the figure> 48 cm² + 80 cm ²
Area of the figure> 128 cm²
Therefore, the area of the figure is 128 cm².
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