Question
In a parallelogram $A B C D$, it is being given that $A B=10\ cm$ and the altitudes corresponding to the sides $A B$ and $A D$ are $D L=6\ cm$ and $B M=8\ cm$, respectively. Find $A D$.
✓
Answer
Since $ABCD$ is a parallelogram and $DL$ is perpendicular to $AB .$
So, its area $=A B \times D L=(10 \times 6)\ cm ^2=60\ cm^2$
Also, in parallelogram $A B C D, B M \perp A D$
$\therefore$ Area of parallelogram $A B C D=A D \times B M 60=A D \times 8\ cm$
$\therefore A D \times 8=60$
$\Rightarrow AD=\frac{60}{8}=7.5\ cm$
$ \therefore AD=7.5\ cm$
So, its area $=A B \times D L=(10 \times 6)\ cm ^2=60\ cm^2$
Also, in parallelogram $A B C D, B M \perp A D$
$\therefore$ Area of parallelogram $A B C D=A D \times B M 60=A D \times 8\ cm$
$\therefore A D \times 8=60$
$\Rightarrow AD=\frac{60}{8}=7.5\ cm$
$ \therefore AD=7.5\ cm$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A triangular park $ABC$ has sides $120 m, 80 m$ and $50 \ m$ . (in a given figure). A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹ $20$ per metre leaving a space $3$ m wide for a gate on one side.
→In the following determine rational numbers $a$ and $b$:
$\frac{4+\sqrt2}{2+\sqrt2}=\text{a}-\sqrt{\text{b}}$
→$\frac{4+\sqrt2}{2+\sqrt2}=\text{a}-\sqrt{\text{b}}$
The lateral surface area of a cube is $900cm^2$. Find its volume.
→Rationalise the denominator: $\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}$
→Simplify: $\Big(\frac{\sqrt{2}}{5}\Big)^8\div\Big(\frac{\sqrt{2}}{5}\Big)^{13}$
→Find six rational numbers between $2$ and $3.$
→If $x = 3$, find the values of the following using in identity: $\Big(\frac{3}{\text{x}}-\frac{\text{x}}{3}\Big)\Big(\frac{\text{x}^2}{9}+\frac{9}{\text{x}^2}+1\Big)$
→Locate $\sqrt{3}$ on the number line.
→Factorize the following expressions:
$8 x^2 y^3-x^5$
→$8 x^2 y^3-x^5$
In figure, lines $AB$ and $CD$ are parallel and $P$ is any point as shown in the figure. Show that $\angle\text{ABP}+\angle\text{CDP}=\angle\text{DPB}.$
→