Download App

Quadrilaterals — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsQuadrilaterals3 Marks
Question
$\text{ABCD}$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$ respectively.

Show that :
$i. \triangle APB \cong \triangle CQD$
$ii. AP = CQ.$
✓

Answer

Given: $\text{ABCD}$ is a parallelogram and $A P$ and $C Q$ are perpendicular from vertices $A$ and $C$ on diagonal $B D$ respectively.
To Prove:
$\triangle APB \cong \triangle CQD$
$A P=C Q$
Proof:
In $\triangle A P B$ and $\triangle C Q D$
$A B=C D \ldots [$Opp. sides of $\| gm \text{ABCD}]$
$\angle A B P=\angle C D Q \ldots [$Alternate interior angles for $A B \| C D]$
$\therefore \angle APB =\angle CQD \ldots\left[\right.$ Each $\left.90^{\circ}\right]$
$\triangle APB \cong \triangle C Q D \ldots [$By $\text{AAS}$ rule$]$
As $\triangle A P B \cong \triangle C Q D \ldots [$As proved above$]$
$\therefore A P=C Q \ldots [\text {c.p.c.t.}]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "3 Marks Question" from Quadrilaterals→Quadrilaterals — all question types→Maths STD 9 question papers→Gujarat Board STD 9 question papers→Gujarat Board question paper generator→

Similar questions

In Figure $\text{ABC}$ is an isosceles triangle in which $AB = AC. AD$ bisects exterior $\angle PAC$ and $CD \| AB$. Show that
$i.\angle DAC = \angle BCA$
$ii. \text{ABCD}$ is a parallelogram
→
In the adjoining figure, $\text{MNPQ}$ and $\text{ABPQ}$ are parallelograms and $T$ is any point on the side $BP.$ Prove that
$i. \text{ar}(\text{MNPQ})=\text{ar}(\text{ABPQ})$
$ii.  \text{ar}(\triangle\text{ATQ})=\frac{1}{2}\text{ar}\text{(MNPQ}).$

→
Find the lateral surface area and total surface area of a cube of edge $10\ cm.$
→
Find the following products: $(2a - 3b - 2c)(4a^2 + 9b^2 + 4c^2 + 6ab - 6bc + 4ca)$
→
A conical tent is $10\ m$ high and the radius of it base is $24\ m$. Find the slant height of the tent. If the cost of $1m^2$ canvas is $Rs. 70$, find the cost of canvas required for the tent.
→
Evaluate: $\frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}$
→
The capacity of a closed cylindrical vessel of height $1m$ is $15.4$ liters. How many square meters of the metal sheet would be needed to make it?
→
Find the volume of a sphere whose surface area is $154cm2.$
→
Find six rational numbers between $2$ and $3.$
→
$P$ is any point on the diagonal $AC$ of a parallelogram $ABCD.$ Prove that $\text{ar}(\triangle\text{ADP})=\text{ar}(\triangle\text{ABP}).$
→