Question

XD is a median of triangle XYZ Eis a midpoint XD such that XE=ED find ar of triangle yxd ar xez

Answer 1

$i ) In \: \triangle XYZ ;\: XD \:is \:a \: mid-point$

$\therefore area(\triangle YXD) \\= Area(\triangle XDZ)\\= \frac{1}{2} \times area ( \triangle XYZ ) \: ---(1)$

$\pink { ( Median \:of \:a \: triangle \:divides \:it }\\\pink { into \:two \:triangles \:of \:equal \:areas )}$

$ii ) area(\triangle XEZ) \\= \frac{1}{2} \times area ( \triangle XDZ ) \: \blue { ( E \:is \: mid-point \:of \:XD )}\\= \frac{1}{2} \times \frac{1}{2} \times area ( \triangle XYZ )\: [ From \: (1) ] \\= \frac{1}{4} \times area ( \triangle XYZ )$

Therefore.,

$\red{i) area(\triangle YXD)} \\\green {= Area(\triangle XDZ)}\\\green {= \frac{1}{2} \times area ( \triangle XYZ ) }$

$\red{ii ) area(\triangle XEZ)}\\\green {= \frac{1}{4} \times area ( \triangle XYZ )}$

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Answer 2

Step-by-step explanation:

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