Question
Construct a quadrilateral $\text{ABCD},$ when:$\angle A = 90^o, AB = 4.6\ cm, BD = 6.4\ cm, AC = 6.0\ cm$ and $CD = 4.2\ cm.$
✓
Answer
Steps:
$1.$ Draw $AB=4.6\ cm$.
$2.$ Through $A$, draw $AP$ such that Angle $A=90^{\circ}$.
$3.$ With $B$ as a center and radii $6.4 \ cm$ draw an arc at $D$ on $AP.$
$4.$ With $D$ and $A$ as a center and radii, $4.2 \ cm$ and $6 \ cm$ draw an arc cutting each other at $C$.
$5.$ Now join $BD , AC$, and $CB$.
$\text{ABCD}$ is the required quadrilateral.
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
The total surface area of a cube is $864 \ cm^2$. Find its volume.
→If $\log _{10} 2=a$ and $\log _{10} 3=b ;$ express each of the following in terms of $'a\ '$ and $'b\ ' :\log 3 \frac{1}{8}$
→If $2^x=3^y=12^z$, show that $: \frac{1}{z}=\frac{1}{y}+\frac{2}{x}$.
[ Hint. Let $2^x=3^y=12^z=k$. Then, $2=k^{\frac{1}{x}}, 3=k^{\frac{1}{y}}$ and $12=k^{\frac{1}{z}}$.
Now, $\left.12=2^2 \times 3 \Rightarrow k^{\frac{1}{2}}=\left(k^{\frac{1}{x}}\right)^2 \times\left(k^{\frac{1}{y}}\right) \Rightarrow k^{\frac{1}{2}}=k^{\left(\frac{2}{x}+\frac{1}{y}\right)}\right]$
→[ Hint. Let $2^x=3^y=12^z=k$. Then, $2=k^{\frac{1}{x}}, 3=k^{\frac{1}{y}}$ and $12=k^{\frac{1}{z}}$.
Now, $\left.12=2^2 \times 3 \Rightarrow k^{\frac{1}{2}}=\left(k^{\frac{1}{x}}\right)^2 \times\left(k^{\frac{1}{y}}\right) \Rightarrow k^{\frac{1}{2}}=k^{\left(\frac{2}{x}+\frac{1}{y}\right)}\right]$
Calculate the amount and the compound interest on ₹ 16000 for 1 year at 15% per annum, compounded half-yearly.
If $A = 30^\circ ;$show that$:4 \cos A \cos (60^\circ - A). \cos (60^\circ + A) = \cos 3A$
→A rectangle $\text{PQRS}$ is drawn on the coordinate axes such that $P$ is the origin, $PQ = 6$ units and $PS = 5$ units. Find the coordinates of the vertices $P, Q R$ and $S.$ Also, find the area of the rectangle.
→Solve for $x :p ^{-5}=\frac{1}{ p ^{x+1}}$
→D is a point on base BC of a $\triangle ABC$ such that $2 BD = DC$.
Prove that : $\operatorname{ar}(\triangle ABD )=\frac{1}{3}$ ar $(\triangle ABC )$.
→Prove that : $\operatorname{ar}(\triangle ABD )=\frac{1}{3}$ ar $(\triangle ABC )$.
If $\sin\theta = \cos\theta $ and $0^\circ < \theta <90^\circ $, find the value of $'\theta '.$
→Find three rational numbers between:$-\frac{1}{2}$ and $\frac{1}{3}$
→