3 Marks Question

Question 13 Marks

If $P$ is any point in the interior of a parallelogram $ABCD$, then prove that area of the triangle $APB$ is less than half the area of parallelogram.

Answer

Draw $\text{DN}\perp\text{AB}$ and $\text{PM}\perp\text{AB}$
Now, ar ($||^{gm}$ $ABCD$)
$=\text{AB}\times\text{DN},\ \text{ar}(\triangle\text{APB})=\Big(\frac{1}{2}\Big)(\text{AB}\times\text{PM})$
Now, $PM < DN$
$\Rightarrow AB \times PM < AB \times DN$
$\Rightarrow\Big(\frac{1}{2}\Big)(\text{AB}\times\text{PM})<\Big(\frac{1}{2}\Big)(\text{AB}\times\text{DN})$
$\Rightarrow\text{ar}(\triangle\text{APB})<\frac{1}{2}\text{ar}(||^{\text{gm}}\text{ABCD})$

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Question 23 Marks

$ABCD$ is a parallelogram whose diagonals $AC$ and $BD$ intersect at $O$. A line through $O$ intersects $AB$ at $P$ and $DC$ at $Q$. Prove that $\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$

Answer

In triangles $POA$ and $QOC$, we have $\angle\text{AOP}=\angle\text{COQ}$
$\text{AO}=\text{OC}$
$\angle\text{PAC}=\angle\text{QCA}$
So, by $ASA$ congruence criterion,
we have $\triangle\text{POA}\cong\triangle\text{QOC}$
$\Rightarrow\text{ar}(\triangle\text{POA})=\text{ar}(\triangle\text{QOC}).$

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Question 33 Marks

Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that: $\text{ar}(\triangle\text{APB})\times\text{ar}(\triangle\text{CPD})\\=\text{ar}(\triangle\text{APD})\times\text{ar}(\triangle\text{BPC})$

Answer

Construction: - Draw $\text{BQ}\perp\text{AC}$ and $\text{DR}\perp\text{AC}$
Proof:- $\text{L.H.S}$ $=\text{ar}(\triangle\text{APB})\times\text{ar}(\triangle\text{CDP})$
$=\Big(\frac{1}{2}\Big)\big[(\text{AP}\times\text{BQ})\big]\times\Big(\frac{1}{2}\Big)\times\text{PC}\times\text{DR}$
$=\Big(\frac{1}{2}\times\text{PC}\times\text{BQ}\Big)\times\Big(\frac{1}{2}\times\text{AP}\times\text{DR}\Big)$
$=\text{ar}(\triangle\text{APD})\times\text{ar}(\triangle\text{BPC}).$
$\text{= R.H.S}$ Hence proved.

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Question 43 Marks

In figure, $OCDE$ is a rectangle inscribed in a quadrant of a circle of radius $10\ cm$. If $\text{OE}=2\sqrt5\text{cm},$ find the area of the rectangle.

Answer

Given $OD = 10\ cm$ and $\text{OE}=2\sqrt5\text{cm}$ By using Pythagoras theorem
$\therefore OD^2 = OE^2 + DE^2$
$\Rightarrow\text{DE}=\sqrt{\text{OD}^2-\text{OE}^2}$
$=\sqrt{10^2-\big(2\sqrt5\big)^2}=4\sqrt5\text{cm}$
$\therefore$ Area of rectangle $OCDE = OE \times DE$ $=2\sqrt5\times4\sqrt5\text{cm}^2=40\text{cm}^2$

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