3 Marks Question

Question 13 Marks

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig.) Prove that ar (LZY) = ar (MZYX)

Answer

We have to prove that ar (LZY) = ar (MZYX)
Since $\triangle\text{LXZ}$ and $(\triangle\text{XMZ})$ are on the same base XZ and between the same parallels LM and XZ, we have
$\triangle(\text{LXZ})=\text{ar}(\triangle\text{XMZ})\ ...(1)$
Adding $\text{ar}(\triangle\text{XYZ})$ to both sides of (1), we get
$\text{ar}(\triangle\text{LXZ})+\text{ar}(\triangle\text{XYZ})=\text{ar}(\triangle\text{XMZ})+\text{ar}(\triangle\text{XYZ})$
$\Rightarrow\text{ar}(\triangle\text{LZY})=\text{ar}(\text{MZTX})$

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Question 23 Marks

ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF, prove th at $\text{ar}(\triangle\text{AER})= \text{ar}(\triangle\text{AFR}).$

Answer

Given In square ABCD, E and F are the mid point of BC and CD respectively. Also R is the mid point of EF.To prove: ar (AER) = ar (AFR)
Construction: Draw $\text{AN}\perp\text{EF}.$
Proof: $\text{ar}(\triangle\text{AER})=\frac{1}{2}\times\text{Base}\times\text{Height}$
$=\frac{1}{2}\times\text{ER}\times\text{AN}$
$=\frac{1}{2}\times\text{FR}\times\text{AN}$ [$\because$ R is the mid - point of EF, so ER = FR]
$=\text{ar}(\triangle\text{AFR})$

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