GSEB Class 7 Maths Solutions Chapter 10 પ્રાયોગિક ભૂમિતિ Exercise 10.5

Gujarat Board Textbook Solutions Class 7 Maths Chapter 10 પ્રાયોગિક ભૂમિતિ Ex 10.5

Question 1. \( m\angle \mathbf{Q} = 90^\circ \), QR = 8 સેમી અને PR = 10 સેમી હોય તેવો કાટકોણ APQR રચો.
Answer:
Here are the construction steps to make the right-angled triangle PQR:
1. First, **make** a line segment QR that **is** 8 cm long. 2. Next, at the point Q, **make** a ray QX that **goes straight up** from QR, forming an angle of \( 90^\circ \). 3. Then, using R as the center, **draw** a curved line (an arc) with a **measurement** of 10 cm. This arc **will cut** the ray QX at a new point, which we **name** P. 4. Finally, **connect** the points P and R **with a line**. 5. The triangle PQR **that you have built** is the right-angled triangle **asked for**.

In simple words: First, draw the base line QR. At Q, draw a right angle. From R, draw an arc with the hypotenuse length (10 cm) to find point P on the right angle line. Join P and R to finish the triangle.

Exam Tip: Always begin by drawing a rough sketch to help visualize the triangle and its given measurements before starting the actual construction.

Question 2. એવો કાટકોણ ત્રિકોણ રચો કે જેના કર્ણની લંબાઈ 6 સેમી અને એક બાજુની લંબાઈ 4 સેમી હોય.
Answer:
Let's construct a right-angled triangle ABC where \( m\angle B = 90^\circ \), the hypotenuse AC is 6 cm, and one side BC is 4 cm.
1. First, **make** a line segment BC that **is** 4 cm long. 2. Next, at the point B, **make** a ray BX that **goes straight up** from BC, forming an angle of \( 90^\circ \). 3. Then, using C as the center, **draw** a curved line (an arc) with a **measurement** of 6 cm. This arc **will cut** the ray BX at a new point, which we **name** A. 4. Finally, **connect** the points A and C **with a line**. 5. The triangle ABC **that you have built** is the right-angled triangle **asked for**.

In simple words: Start by drawing the given side. At one end of this side, construct a right angle. From the other end of the side, use the hypotenuse length to draw an arc, finding the third vertex. Join the points to complete the triangle.

Exam Tip: Remember that in a right-angled triangle, the hypotenuse is always the longest side, and it's opposite the right angle.

Question 3. સમઢિબાજુ કાટકોણ ત્રિકોણ ABC રચો, જેમાં \( m\angle \mathbf{ACB} = 90^\circ \) અને AC = 6 સેમી છે.
Answer:
Let's construct an isosceles right-angled triangle ABC, where \( m\angle ACB = 90^\circ \) and AC = 6 cm. Since it's an isosceles right-angled triangle, the two perpendicular sides must be equal, so BC will also be 6 cm.
1. First, **make** a line segment AC that **is** 6 cm long. 2. Next, at the point C, **make** a ray CX that **goes straight up** from AC, forming an angle of \( 90^\circ \). 3. Then, using C as the center, **draw** a curved line (an arc) with a **measurement** of 6 cm. This arc **will cut** the ray CX at a new point, which we **name** B. 4. Finally, **connect** the points A and B **with a line**. 5. The triangle ABC **that you have built** is the right-angled triangle **asked for**.

In simple words: Draw one equal side, then make a 90-degree angle at the common vertex. Measure the second equal side along this angle. Connect the endpoints to form the isosceles right-angled triangle.

Exam Tip: For an isosceles right-angled triangle, remember that the two sides forming the right angle are always equal in length.