Class Vii Math Solution WBBSE Geometry Chapter 4 Construction Of Triangles Exercise 4 Solved Problems
Question 1. To construct an angle equal to the given angle.
Solution:
Given ABC is an angle of measure x°
Method:
1. I draw a straight line QR. With B as a centre and with any radius an arc is drawn to intersect the AB and BC at D and E respectively.
2. With Q as a centre and with the same radius another arc is drawn to cut the straight line QR at M.
3. Now with M as a centre and with a radius equal to DE another arc is drawn to cut the previous arc at N. In joined Q, N and produced to P.
∠PQR is the required angle where ∠PQR = ∠ABC = x°
Wbbse Class 7 Maths Solutions
Question 2. Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 5 cm
[ For any triangle the sum of lengths of its two smaller sides must be greater than the third side]
Solution:
Method:
1. At first, using scale and pencil we draw three line segments of length 4 cm, 5 cm and 6 cm.
2. I draw a ray BX with centre B and radius 6 cm an arc is drawn which cut BX at C.
3. With B and C as centres and with respective radius 4 cm and 5 cm two arcs are drawn on the same side of BC such that these two arcs intersect each other at A. A, B and A, C are joined.
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∴ ABC is the required triangle whose AB = 4 cm, BC = 6 cm and AC = 5 cm.
Question 3. Construct a triangle ABC in which AB = 4.6 cm, BC= 6.2 cm and ∠ABC = 70°
Solution:
Method :
1. At first, I take two line segments of length 4.6 cm and 6.2 cm using scale and pencil and an angle of 70° with the protractor.
2. A rays BX is taken. At B on BX, ∠YBX = 70° is drawn. From BX, the portion BC = 6.2 cm is cut off and from BY, the portion BA = 4.6 cm is cut off. A, C is joined.
∴ ABC is required triangle whose AB = 4.6 cm, BC = 6.2 cm and ∠ABC = 70°
Wbbse Class 7 Maths Solutions
Question 4. Draw a triangle PQR whose QR
Solution:
Method:
1. I take a ray QX and cut off a line segment QR of length 7.5 cm
2. At Q and R two angles ∠YQX and ∠ZRQ are drawn where ∠YQX = 60° and ∠ZRQ = 45°. QY and RZ intersect at P.
So PQR is the required triangle whose QR = 7.5 cm, ∠PQR = 60° and ∠PRQ = 45°.
Wbbse Class 7 Maths Solutions
Question 5. Draw a right-angle triangle ABC such that ∠ABC = 90° and AC = 9.6 cm.
Solution:
Method:
1. I draw line segment AB = 5 cm and AC = 9.6 cm
2. A ray BX is drawn. At B on BX, ∠YBX = 90° is drawn.
3. I take a line segment BA = 5 cm.
4. With A as a centre and with a radius equal to 9.6 cm an arc is drawn which intersects BX at C. I join A, C. So ΔABC is the required triangle.
Construction Of Triangles
Class Vii Math Solution WBBSE Construction Of Triangles Exercise 8.1
1. The length of the three sides of the triangles is given. Let’s identify the cases where, triangles can be constructed and construct those triangles, and give reasons for the cases where the triangle cannot be constructed.
- 4cm, 5cm & 7cm.
- 9cm, 4cm. And 4cm.
- 6cm, 8cm and 10cm.
Solution :
Given
The length of the three sides of the triangles is given.
4cm, 5cm & 7cm.
For the construction of any triangle when the length of 3 sides is given, the sum of the length of two smaller sides must be greater than the third side.
In the 1st case. Here 4 cm + 5cm > 7cm
∴ It is possible to construct a triangle
In the 2nd case, here 4cm + 4cm < 9cm.
∴ It is not possible to construct a triangle
In the 3rd case, here 6cm + 8cm > 1 0 cm
∴ It is possible to construct a triangle
2. Let’s draw a triangle ABC in which AB = 5.5cm, BC = 5cm and CA= 6cm.
Solution :
1. At first, using scale & pencil we draw three line segments of length 5cm, 6cm & 5.5cm.
2. Let’s draw a ray AX, with centre A & radius 6cm draw an arc on AX, such that AC = 6cm.
3. With centres A & C and with radius 5.5cm & 5cm respectively draw two arcs which cut each other’s B.
Points A, B and B, C are joined with scale to get the required triangle ABC, in which AB = 5,5cm, BC = 5cm & AC = 6cm.
3. Let’s draw an equilateral triangle having each side = 4.5cm. Let’s also measure its three angles with a protractor and write their measures.
Solution :
From the line BX, cut a line segment BC = 4.5cm
Now with centres B & C with a radius of 4.5cm, draw two arcs. Which cut each other at A. By joining AB & AC, we get the required equilateral triangle
ABC. By measuring the angles with the protector we get that each angle = 60°.
4. Let’s draw a triangle PQR, such that PQ = 6m, QR = 5cm, and PR = 6cm. Let’s measure each of its angles with a protractor and write them down.
Solution :
From the line QX, cut a line segment QR = 5cm with centre Q & R with radius 6cm two arcs are drawn which cut each other at P.
By joining PR & PQ, we get an isosceles triangle PQR, whose ∠ PQR = ∠ PRQ = 65° and∠ QPR = 50°
WBBSE Class 7 Math Solution Construction Of Triangles Exercise 8.2
1. Let’s construct a triangle ABC in which AB = 4cm, BC = 6cm and ∠ ABC = 45°
Solution :
Given
First, take a ray BX. From BX cut BC = 6cm.
Now at B, draw an angle ∠ PBC = 45° with compass & scale. From BP cuts BA is equal to 4cm. Join AC.
∴ ABC is the required triangle whose AB = 4cm
BC = 6cm & included angle ABC = 45°
3. Let’s draw a triangle PQR, such that PQ = 4cm, QR = 3cm, and ∠ PQR = 90° in Δ PQR, let’s also measure the side PR with scale and take down the measured value.
Solution :
Given
Take a ray QX, from QX, cut QR = 3cm. Now draw ∠ YQR.= 90 from QY cut QP = 4cm, join PR.
∴ PQR is the required right-angled triangle whose PQ = 4cm, QR = 3cm & ∠ PQR = 90°
Now the length of PR = 5cm by measuring with scale.
4. Let’s draw an isosceles triangle whose two equal sides are 7.2 cm each and the angle included between them is – 100°
Solution :
Given
From the ray BX, cut off BC = 7.2cm.
Now at B draw an angle YBC = 100° (given) with a compass.
From BY, cut BA = 7.2cm, and Join AC.
∴ ABC is the required isosceles triangle whose two equal sides BC
= AB = 7.2cm & the included angle ABC = 100°
WBBSE Class 7 Math Solution Construction Of Triangles Exercise 8.3
1. Let’s draw a triangle Δ XYZ, such that YZ – 6.5 cm. ∠ XYZ = 60 and ∠ XZY = 70°
Solution :
Given
From the ray -YR cuts YZ = 6.5 cm
At Y draw an angle PYZ = 60° & at Z draw an angle QZY = 70°
Let YP & ZQ cut each other at X
∴ XYZ is the required triangle whose YZ = 6.5 XYZ = 603 & ∠ XZY = 70°
3. Let’s construct a Δ ABC, so that BC = 5.5 cm, ∠ ABC = 60° and ∠ ACB = 30
Solution :
Given
From the ray BX, cut BC = 5.5cm. Now with the compass draw an angle at B,
MBC = 60° & at C draw an angle NCB = 30°.
MB & NC cut each other at A.
∴ ABC is the required triangle whose BC = 5.5cm &
∠ ABC = 60° ∠ ACB = 30°
4. Let’s try to draw a triangle whose one side QR = 7.2cm, ∠ PQR = 80° and ∠ PRQ = 115°. If a triangle can not be constructed, let’s try to find the reason for such a problem.
Solution :
In this case triangle cannot be constructed as here some of the two angles is 80° + 11 5° = 1 95°. More than 1 80°
We know that the sum of the 3 angles must be equal to 180°
5. Let’s draw an isosceles triangle Δ DEF in which the side EF is of length 6.2 cm and the two angles, adjacent to the side have their sum = 100°
Solution:
Given
From the ray, EX cut EF = 6.2cm. Now at E & F draw two angles PEF & QFE, both equal to 50°
As here the sum of the two equal angles = 100° (given) as the triangle is an isosceles Δ PE & QF in interseptal Δ DEF is the required isosceles triangle.
WBBSE Class 7 Math Solution Construction Of Triangles Exercise 8.4
1. Let’s construct a right-angled triangle in Δ PQR such that ∠ PQR = 90°, PQ = 6cm. And QR = 4cm.
Solution :
Given
From the ray QX, QR = 4cm is cut off
At Q, Z YQX = 90° is drawn from YQ cut off PQ = 6cm. Joined PR
Hence, A PQR is the required right-angled triangle.
2. Let’s draw a right-angled isosceles triangle Δ ABC, such that ∠ ABC = 90° & AB = 7cm.
Solution:
To draw an isosceles right-angled triangle ABC
Here Δ ABC is the required isosceles triangle as AB = BC = 7cm & ∠ ABC = 90°.
3. Let’s draw a right-angled triangle Δ XYZ such that ∠ XYZ = 90, XZ = 10cm and XY = 6cm.
Solution:
To draw a right-angled triangle Δ XYZ, such that ∠ XYZ = 90°, XZ = 1 0 cm & XY = 6cm.
XYZ is the required right angle triangle whose ∠ XYZ = 90° & XZ = 10 cm, XY = 6cm
∴ YZ = 8cm.
4. Let’s draw a right-angled triangle Δ ABC. Such that ∠ BAC = 90°, 6cm BC = 8cm and ∠ ACB = 45°.
Solution :
1. Draw a ray CX. Now, at C on CX an angle XCD = 90° is drawn.
2. ∠ XCD is bisected, we get ∠ XCY = 45°
3. From CY, CB = 8cm is cut off. From B, a perpendicular BA is drawn which cuts CX at A.
4. Δ ABC is the required triangle.
Where BC = 8cm, ∠ BAC = 90° & ∠ BCA = 45°
