Question
Draw a triangle $A B C$ with $B C=7 cm, \angle B =45^{\circ}$ and $\angle C =60^{\circ}$. Then construct another triangle, whose sides are $\frac{3}{5}$ times the corresponding sides of $\triangle A B C$.
✓
Answer
In order to construct the given triangle, follow the following steps:
Step 1: Draw $B C=7 cm$.
Step 2: At $B$, construct $\angle B=45^{\circ}$ and at $C$, construct $\angle C=60^{\circ}$. They intersect each other at $A$. Thus, $\triangle A B C$ is constructed.
Step 3: Construct an acute angle $\angle C B Z$ at $B$ on opposite side of vertex $A$ of $\triangle A B C$
Step 4: Along $B Z$, mark off 5 points $B_1, B_2, B_3, B_4$, $B_5$ such that
$
B B_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5 .
$
Step 5: Join $B_5 C$
Step 6: Since we have to construct a triangle each of whose sides is $\frac{3}{5}$ of the corresponding sides of $\triangle A B C$. So, take three parts out of five equal parts on $B Z$ i.e., from $B_3$, draw $B_3 C| | B_5 C$, meeting $B C B C$ at $C$.
Step 7: From $C^{\prime}$, draw $C^{\prime} A^{\prime} \mid I C A$, meeting $B A$ at $A^{\prime}$. Thus, $A^{\prime} B C^{\prime}$ is the required triangle, each of whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle A B C$ as shown below
Step 1: Draw $B C=7 cm$.
Step 2: At $B$, construct $\angle B=45^{\circ}$ and at $C$, construct $\angle C=60^{\circ}$. They intersect each other at $A$. Thus, $\triangle A B C$ is constructed.
Step 3: Construct an acute angle $\angle C B Z$ at $B$ on opposite side of vertex $A$ of $\triangle A B C$
Step 4: Along $B Z$, mark off 5 points $B_1, B_2, B_3, B_4$, $B_5$ such that
$
B B_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5 .
$
Step 5: Join $B_5 C$
Step 6: Since we have to construct a triangle each of whose sides is $\frac{3}{5}$ of the corresponding sides of $\triangle A B C$. So, take three parts out of five equal parts on $B Z$ i.e., from $B_3$, draw $B_3 C| | B_5 C$, meeting $B C B C$ at $C$.
Step 7: From $C^{\prime}$, draw $C^{\prime} A^{\prime} \mid I C A$, meeting $B A$ at $A^{\prime}$. Thus, $A^{\prime} B C^{\prime}$ is the required triangle, each of whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle A B C$ as shown below
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 racetrack is in the form of a ring whose inner circumfarence is 352m and outer circumference is 396m. Find the width and the area of the track. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→A lawn is in the from of a rectangle whose sides are in the ratio 5 : 3 and its area is $3375m^2$. Find the cost of fencing the lawn at ₹ 20 per meter.
→The length of three concesutive sides of a quadilateral circumscribing a circle are 4cm, 5cm, 7cm respectively. Determine the length of thefourth side.
A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is:
- Red or white.
- Not black.
- Neither white nor black.
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
$\sin\theta=\frac{11}{15}$
→$\sin\theta=\frac{11}{15}$
O is any point inside a rectangle ABCD see Fig. Prove that $OB^2 + OD^2 = OA^2 + OC^2$
.
→.
Prove that following numbers are irrationals:
$\frac{3}{2\sqrt{5}}$
→$\frac{3}{2\sqrt{5}}$
Find the area of $\triangle\text{ABC}$ whose vertices are:
A(-5, 2), B(-4, -5) and C(4, 5)
→A(-5, 2), B(-4, -5) and C(4, 5)
Find the lengths of the arcs cut off from a circle of radius 12cm by a chord 12cm long. Also, find the area of the minor gment. $\big[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.73.\big]$
→The height of a tower is 10m. What is the length of its shadow when Sun's altitude is 45°?