GSEB Class 7 Maths Solutions Chapter 10 Practical Geometry Exercise 10.5

Gujarat Board Textbook Solutions Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1. Construct the right-angled \( \triangle PQR \), where \( m\angle Q = 90^\circ \), QR = 8 cm and PR = 10 cm.
Answer: Here are the steps for its construction:
I. First, draw a line segment QR which is 8 cm long.
II. At point Q, build \( \angle RQX \) so that its measure is \( 90^\circ \).
III. With R as the center, and a radius of 10 cm, draw an arc that cuts the ray \( \overrightarrow{QX} \) at point P.
IV. Connect points P and R.
Consequently, \( \triangle PQR \) is the right-angled triangle that was asked for.

Exam Tip: When constructing a right-angled triangle with a given hypotenuse, always remember to use a compass to mark the hypotenuse length from the non-right-angle vertex to cut the perpendicular line.

Question 2. Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Answer: Here are the steps for its construction:
I. Start by drawing a line segment MN, which is 4 cm in length.
II. At point M, build \( \angle NMX \) with a measure of \( 90^\circ \).
III. Using N as the center, and a radius of 6 cm (the hypotenuse length), draw an arc that cuts the ray \( \overrightarrow{MX} \) at point L.
IV. Connect points L and N.
Therefore, \( \triangle LMN \) is the right-angled triangle that was needed.

Exam Tip: When given a hypotenuse and one leg, always draw the given leg first, construct the right angle, and then use the compass to cut the perpendicular at the hypotenuse length from the other endpoint of the leg.

Question 3. Construct an isosceles right-angled triangle ABC, where \( m\angle ACB = 90^\circ \) and AC = 6 cm.
Answer: Here are the steps for its construction:
I. First, draw a line segment AC which is 6 cm long.
II. At point C, build \( \angle ACX \) with a measure of \( 90^\circ \).
III. With C as the center, and a radius equal to AC (6 cm), draw an arc that cuts the ray \( \overrightarrow{CX} \) at point B.
IV. Connect points B and A.
Thus, \( \triangle ABC \) is the isosceles right-angled triangle that was required.

Exam Tip: For an isosceles right-angled triangle, remember that the two legs meeting at the right angle must have equal lengths. This simplifies the construction as one leg's length determines the other's.