Question
In figure, compute the area of quadrilateral ABCD.
✓
Answer
Given that: $DC =17 cm A D=9 cm$ and $B C=8 cm$ In
$\triangle BCD$
we have $C D^2=B D^2+B C^2$
$\Rightarrow(17)^2=B D^2+(8)^2$
$\Rightarrow B D^2=289-64$
$\Rightarrow B D=15 \ln$
$\triangle ABD$,
we have $A B^2+A D^2=B D^2$
$\Rightarrow(15)^2=A B^2+(9)^2$
$\Rightarrow A B^2=225-81$
$\Rightarrow A B=123 \operatorname{ar}(\text { quad, } A B C D)$
$=\operatorname{ar}(\triangle ABD)+(\triangle BCD)$
$\Rightarrow \operatorname{ar}(\text { quad, } A B C D)=\frac{1}{2}(12 \times 9)+\frac{1}{2}(8 \times 17)=54+68$
$=112 cm^2$
$\Rightarrow \operatorname{ar}(\text { quad, } A B C D)=\frac{1}{2}(12 \times 9)+\frac{1}{2}(8 \times 17)$
$=54+60 cm^2$
$=114 cm^2$
$\triangle BCD$
we have $C D^2=B D^2+B C^2$
$\Rightarrow(17)^2=B D^2+(8)^2$
$\Rightarrow B D^2=289-64$
$\Rightarrow B D=15 \ln$
$\triangle ABD$,
we have $A B^2+A D^2=B D^2$
$\Rightarrow(15)^2=A B^2+(9)^2$
$\Rightarrow A B^2=225-81$
$\Rightarrow A B=123 \operatorname{ar}(\text { quad, } A B C D)$
$=\operatorname{ar}(\triangle ABD)+(\triangle BCD)$
$\Rightarrow \operatorname{ar}(\text { quad, } A B C D)=\frac{1}{2}(12 \times 9)+\frac{1}{2}(8 \times 17)=54+68$
$=112 cm^2$
$\Rightarrow \operatorname{ar}(\text { quad, } A B C D)=\frac{1}{2}(12 \times 9)+\frac{1}{2}(8 \times 17)$
$=54+60 cm^2$
$=114 cm^2$
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