Question

$KLMN$ is an isosceles trapezium whose diagonals cut at $X$ and $KL$ is parallel to $NM.$ If $\angle KNL={25}^{\circ },\angle KMN={30}^{\circ }$, find $\left(a\right)$ $\angle KXN\left(b\right)\angle MLN$.

• A. $\angle KXN={60}^{\circ }$
• B. $\angle KXN={80}^{\circ }$
• C. $\angle MLN={85}^{\circ }$
• D. $\angle MLN={95}^{\circ }$

Answer 1

Answer: (A), (D)

$\angle MLN={95}^{\circ }$
$\angle KXN={60}^{\circ }$

In $\Delta s$ $KLN$ and $KLM$
$KN=LM$ (Isos. trap)
$\angle NKL=\angle KLM$ (Prop. of isos. trap.)
$KL=KL$ (Common side)
$\therefore \Delta KLN\cong \Delta KLM$ ....SAS

$\Rightarrow \angle KNL=\angle KML$ ....cpct
Thus $\angle LMN={25}^{\circ }+{30}^{\circ }={55}^{\circ }$
Now, $\angle KNM=\angle LMN$
$\Rightarrow KNL+\angle LNM={55}^{\circ }\Rightarrow {55}^{\circ }-{25}^{\circ }={35}^{\circ }$
In $\Delta NXM,\angle NXM={180}^{\circ }-(\angle XNM+\angle XMN)$
$={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
(i) Now, $\angle KXN={180}^{\circ }-\angle NXM={180}^{\circ }-{120}^{\circ }={60}^{\circ }$ (Linear pair)
Also, $\angle LXM=\angle KXN={60}^{\circ }$
(ii) Now, in $\Delta MLX,\angle MLX\left(\angle MLN\right)={180}^{\circ }-({60}^{\circ }+{25}^{\circ })$
$={180}^{\circ }-{85}^{\circ }={95}^{\circ }$