Chapter 10 Congruent Triangles RD Sharma Solutions Exercise 10.1 Class 9 Maths
Chapter Name | RD Sharma Chapter 10 Congruent Triangles Exercise 10.1 |
Book Name | RD Sharma Mathematics for Class 10 |
Other Exercises | - Exercise 10.2
- Exercise 10.3
- Exercise 10.4
- Exercise 10.5
- Exercise 10.6
|
Related Study | NCERT Solutions for Class 10 Maths |
Exercise 10.1 Solutions
1. In Fig. 10.22, the sides BA and CA have been produced such that : BA = AD and CA = AE. Prove that segment DE || BC.
Solution
Given that, the sides BA and CA have been produced such that BA = AD and CA = AE and given to prove DE || BC
Consider triangle BAC and DAE,
We have
BA = AD and CA = AE [∵ given in the data]
And also ∠BAC = ∠DAE [∵ vertically opposite angles]
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgyVPeWZbkHzLnlmVIPa4YdgAn0gHxb6i3RncU9-_YnFg0yBrZEcmOrbnvyx6H0iHWQSeQTiVgbRIM-CIarc3gNeZjgZ1titzpNGsQzpQrlOTx6fv5ziuAdvQ8Vrty7KJiEiz9DbJVq5J3JT1Q1J1BVYZX8lpk2r1qcxNhflTx2dCdqh7_ZUvz9p3NS/w149-h157/Class%209%20Chapter%20-%2010%20Congruent%20Triangles%20%20Exercise%20-%2010.1%20img%202.JPG)
So, by SAS congruence criterion, we have ΔBAC ≅ ΔDAE
⇒ BC = DE and ∠DEA = ∠BCA, ∠EDA = ∠CBA
[Corresponding parts of congruent triangles are equal]
Now, DE and BC are two lines intersected by a transversal DB such that ∠DEA = ∠BCA,
i.e., alternate angles are equal
Therefore, DE || BC
2. In a ΔPQR, if PQ = QR and L, M and N are the mid - points of the sides PQ, QR and RP respectively. Prove that : LN = MN.
Solution
Given that, in ΔPQR, PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP respectively and given to prove that LN = MN
Here we can observe that PQR is and isosceles triangle
⇒ PQ = QR and ∠QPR = ∠QRP ...(1)
And also, L and M are midpoints of PQ and QR respectively
⇒ PL = LQ = PQ/2 , QM = MR = QR/2
And also, PQ = QR
⇒ PL = LQ = QM = MR = PQ/2 = QR/2 ...(2)
Now, consider ΔLPN and ΔMRN,
LP = MR [From (2)]
∠LPN = ∠MRN [From (1)]
∵∠QPR and ∠LPN and ∠QRP and ∠MRN are same
PN = NR [∵ N is midpoint of PR]
So, by SAS congruence criterion, we have ΔLPN = ΔMRN
⇒ LN = MN
[∵ Corresponding parts of congruent triangles are equal]
3. In Fig. 10.23, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiNDxIUY7XcLhXrmKoxXVsFHf09eZsBFW6dv4EySzVtU5P2jsDfBl4kYIuK9EsQXYxDhyL_EdJ5dK5OYkoxYXV7rCg-b3jMDw-vK6YU2z6kq_uN5F1rNc8JHuPIxLjGFmwXxVGfOzFtVJKVwCHTvrIsRJYK9lY0gXnuhupRjtK4JJIvmx_02WR6gqoP/w115-h186/Class%209%20Chapter%20-%2010%20Congruent%20Triangles%20%20Exercise%20-%2010.1%20img%203.JPG)
Solution
Given that PQRS is a square and SRT is an equilateral triangle. And given to prove that ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwBaIk4H-5QPddYM9dPSc8nL_UGb0kF3MMZiVkBmwlu1RrEKRDLuvhgdGgjXFtOuOS1BFhIBGY4B4fFFgKam93QQsZdFU6HJZlKnPZqRQlOJ_awcX9mrGSi1MNZWU-oOWfUciS69e1a5Ex3ccGjiIsmgVwGgWaXBtW2_R3czFmPIxb_6zLhwlGcxRy/w572-h724/Class%209%20Chapter%20-%2010%20Congruent%20Triangles%20%20Exercise%20-%2010.1%20img%204.JPG)
4. Prove that the medians of an equilateral triangle are equal. Solution
Given to prove that the medians of an equilateral triangle are equal
Median : The line joining the vertex and midpoint o opposite side.
Now, consider an equilateral triangle ABC
Let D, E, F are midpoints of BC, CA and AB.
Then, AD, BE and CF are medians of ΔABC.
Now,
D is midpoint of BC ⇒ BD = DC = BC/2
Similarly, CE = EA = AC/2
AF = FB = AB/2
Since, ΔABC is an equilateral triangle
⇒ AB = BC = CA ...(1)
⇒ BD = DC = CE = EA = AF = FB = BC/2 = AC/2 = AB/2 ...(2)
And also, ∠ABC = ∠BCA = ∠CAB = 60°
Now, consider ΔABD and ΔBCE
AB = BC [From (1)]
BD = CE [From (2)]
5. In a ΔABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Solution
Consider a ΔABC
Given that ∠A = 120 and AB = AC and given to find ∠B and ∠C
We can observe that ΔABC is an isosceles triangle since AB = AC
⇒ ∠B = ∠C ...(1)
[Angles opposite to equal sides are equal]
We know that sum of angles in a triangle is equal to 180°
6. In a ΔABC, if AB = AC and ∠B = 70° , find ∠A.
Solution
Consider ΔABC, we have ∠B = 70° and AB = AC
Since, AB = AC ΔABC is an isosceles triangle
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKL5VnUdhHy-eDPO5T7Z7QQPugpLi2dwUnt-lLQbIFLvLI6nJkkg-NnQe0F9x-3318NJeHTLaNoCijytqUYDS7buIqUmpxSfCMjwgcToNiPyO9jYciLvetHrFN4uUAnrIMuy2jWrGR_5-ETmLJ0IB6jZwpLgaX7orfSGwSglImxcr51Xe1MMokkzx6/w599-h239/Class%209%20Chapter%20-%2010%20Congruent%20Triangles%20%20Exercise%20-%2010.1%20img%207.JPG)
7. The vertical angle of an isosceles triangle is 100 .Find its base angles.
Solution
Consider an isosceles ΔABC such that AB = AC
Given that vertical angle A is 100 . Given to find the base angles
Since ΔABC is isosceles
∠B = ∠C [Angles opposite to equal angles are equal]
8. In Fig, 10.24, AB = AC and ∠ACD = 105, find ∠BAC.
Solution
Consider the given figure
We have,
AB = AC and ∠ACD = 105
9. Find the measure of each exterior angle of an equilateral triangle .
Solution
Given to find the measure of each exterior angle of an equilateral triangle consider an equilateral triangle ABC.
We know that for an equilateral triangle
Similarly, we can find ∠FAB and ∠FBC also as 120° because ABC is an equilateral triangle
∴ ∠ACD = ∠EAB = ∠FBC = 120°
Hence, the median of each exterior angle of an equilateral triangle is 120°
10. If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Solution
ED is a straight line segment and B and C are points on it.
11. In Fig. 10.25, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.
Solution
Consider the figure
Given
AB = AC, DB = DC and given to find the ratio
∠ABD = ∠ACD
Now, ΔABC and ΔDBC are isosceles triangles since AB = AC and
DB = DC respectively
⇒ ∠ABC = ∠ACB and ∠DBC = ∠DCB [∵ angles opposite to equal sides are equal]
Now consider,
∠ABD : ∠ACD
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhiNl4ZaPh_pWCB85EDpNI4iffNGtE15wPCmj5lZP1hu6ixSvZP1erDF0eFW7SloH1H59CDh2xGNe7Aotn1eNVt_C8KXuKcstTkdGAoXfRGwgLF1f_xifZf0YqGnoCVvwvYiFQWRGRqIWK_JCWZlR40vZslKYAVI5vIz31kY9Bnp9wdcEgihQIRfGEK/w535-h88/Class%209%20Chapter%20-%2010%20Congruent%20Triangles%20%20Exercise%20-%2010.1%20img%2015.JPG)
12. Determine the measure of each of the equal angles of a right - angled isosceles triangle.
OR
ABC is a right - angled triangle in which ∠A = 90 and AB = AC. Find ∠B and ∠C.
Solution
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgKBrjkd6aeiOw9Whs042am5rT_xhU5rCCiynhUr2Gk6L-LOy1pP47-OmqVa4K4z9Gt25K7oHl0r_PevrvYyCPinpHJa-6kQxRdpUBwtf70sB2uamx1E-PHOSrrK0btfYcLc9T_YnWQZDEhqCHBobJ5ACqTldGbeW0hl8x30TS8xoNqRvPDPasaomWB/w138-h141/Class%209%20Chapter%20-%2010%20Congruent%20Triangles%20%20Exercise%20-%2010.1%20img%2016.JPG)
Given to determine the measure of each of the equal angles of right angled isosceles triangle
Consider on a right - angled isosceles triangle ABC such that
∠A = 90° and AB = AC
Since, AB = AC ⇒ ∠C = ∠B ...(1)
[Angles opposite to equal sides are equal]
Now,
Sum of angles in a triangle = 180°
Hence, the measure of each of the equal angles of a right - angled isosceles triangle is 45° .
13. AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. 10.26). Show that the line PQ is perpendicular bisector of AB.
Solution
Consider the figure,
We have
AB is a line segment and P, Q are points on opposite sides of AB such that
AP = BP ...(1)
AQ = BQ ...(2)
We have to prove that PQ is perpendicular bisector of AB.
Now consider ΔPAQ and ΔPBQ,
We have AP = BP [∵ From (1)]
AQ = BQ [∵ From (2)]
And PQ = PQ [Common site]
⇒ΔPAQ ≅ ΔPBQ ...(3) [From SSS congruence]
Now, we can observe that ΔAPB and ΔABQ are isosceles triangles. (From 1 and 2)
⇒ ∠PAB = ∠PBA and ∠QAB = ∠QBA
Now consider ΔPAC and ΔPBC,
C is the point of intersection of AB and PQ.
PA = PB [From (1)]
∠APC = ∠BPC [From (3)]
PC = PC [Common side]
So, from SAS congruency of triangle ΔPAC ≅ ΔPBC
⇒ AC = CB and ∠PCA = ∠PCB ...(4)
[∵ Corresponding parts of congruent triangles are equal]
And also, ACB is line segment
⇒ ∠ACP + ∠BCP = 180°
But ∠ACP = ∠PCB
⇒ ∠ACP = ∠PCB = 90° ...(5)
We have AC = CB
⇒ C is the midpoint of AB
From (4) and (5)
We can conclude that PC is the perpendicular bisector of AB
Since C is a point on the line PQ, we can say that PQ is the perpendicular bisector of AB.