Класс 9 RD Sharma Solutions – Глава 10 Конгруэнтные треугольники – Упражнение 10.1
Вопрос 1. На рисунке стороны ВА и СА составлены так, что: ВА = AD и СА = AE. Докажите, что отрезок DE || ДО Н.Э.
Решение:
Given that:
BA = AD and CA = AE
Prove that DE || BC
So, here we consider triangle BAC and DAE,
BA = AD -(given)
CA= AE -(given)
And ∠BAC = ∠DAE -(vertically opposite angles)
Hence, by the SAS congruence criterion, we have
Δ BAC ≃ ΔDAE
So, we can say that:
∠DEA = ∠BCA, ∠EDA = ∠CBA -(Corresponding parts of congruent triangles are equal)
Now, lines DE and BC are intersected by a transversal DB such that ∠DEA = ∠BCA.
Hence, proved that DE || BC.
Вопрос 2. В Δ QPR, если PQ = QR и L, M и N — середины сторон PQ, QR и RP соответственно. Докажите, что LN = MN.
Решение:
Given that,
In Δ QPR, PQ = QR
Also, L, M, N are midpoints of the sides PQ, QP, and RP.
And we are given to prove that LN = MN.
In Δ QPR, PQ = QR, and L, M, N are midpoints
Hence, Δ QPR is an isosceles triangle.
PQ = QR
∠ QPR = ∠ QRP -(1)
And also, L and M are midpoints of PQ and QR.
Hence, we can say that:
PQ = QR
PL = LQ = QM = MR = PQ/2 = QR/2
Now, we have Δ LPN and Δ MRN,
LP = MR
∠LPN = ∠MRN -(From above)
∠QPR and ∠LPN are the same.
And also ∠QRP and ∠MRN are the same.
PN = NR -(N is the midpoint of PR)
So, by SAS congruence criterion, we can say that ΔLPN = ΔMRN
By corresponding parts of congruent triangles are equal we can say that LN = MN.
Вопрос 3. На приведенном рисунке PQRS — квадрат, а SRT — равносторонний треугольник. Докажите, что (i) PT = QT (ii) ∠TQR = 15°
Решение:
We are given that PQRS is a square and SRT is an equilateral triangle.
And we have to prove that
(i) PT = QT
PQRS is a square and SRT is an equilateral triangle.
Now we see that PQRS is a square
PQ = QR = RS = SP -(1)
And ∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90° = right angle
And also, SRT is an equilateral triangle.
SR = RT = TS -(2)
And ∠TSR = ∠SRT = ∠RTS = 60°
From equation(1) and (2) we can say that,
PQ = QR = SP = SR = RT = TS -(3)
And also for angles we have,
∠TSP = ∠TSR + ∠RSP = 60° + 90° + 150°
∠TRQ = ∠TRS + ∠SRQ = 60° + 90° + 150°
∠TSR = ∠TRQ = 150° -(4)
SP = RQ -(From eq (3))
Hence, by SAS congruence criterion we can say that ΔTSP = ΔTRQ i.e both the triangles are congruent.
PT = QT (Corresponding parts of congruent triangles are equal)
(ii) ∠TQR = 15°
Let’s see ΔTQR.
QR = TR
And also it is given that ΔTQR is an isosceles triangle.
∠QTR = ∠TQR -(Angles opposite to equal sides)
As we know that the sum of angles in a triangle is equal to 180∘
∠QTR + ∠TQR + ∠TRQ = 180°
2∠TQR + 150° = 180°
2∠TQR = 180° – 150°
2∠TQR = 30°
Hence, ∠TQR = 15°
Hence, proved
Вопрос 4. Докажите, что медианы равностороннего треугольника равны.
Решение:
Prove that the medians of an equilateral triangle are equal.
So, let D, E, F are midpoints of BC, CA, and AB. AD, BE and CF are medians of ABC.
Then, AD, BE and CF are medians of ABC.
Now, we can say that
D Is the midpoint of BC
So now we can say that,
BD = DC = BC/2
CE = EA = AC/2
AF = FB = AB/2
Since, it is already given that ΔABC is an equilateral triangle
Therefore
AB = BC = CA
Hence, we can imply that,
BD = DC = CE = EA = AF = FB = BC/2 = AC/2 = AB/2
Also, we can say that ∠ABC = ∠BCA = ∠CAB = 60° -(Angles of equilateral triangle)
Now, consider ΔABD and ΔBCE AB = BC
Here we find that BD = CE
Now, in ΔTSR and ΔTRQ
TS = TR
∠ABD = ∠BCE
So, from SAS congruence criterion, we can conclude that ΔABD = ΔBCE
Now, consider ΔBCE and ΔCAF, BC = CA
∠BCE = ∠CAF (From above)
CE = AF
So, from SAS congruence criterion, we have, ΔBCE = ΔCAF
BE = CF [from above]
AD = BE = CF.
Hence proved
Вопрос 5. В ΔABC, если ∠A = 120° и AB = AC. Найдите ∠B и ∠C.
Решение:
In the triangle ABC
Given that ∠A = 120° and AB = AC
According to the question the triangle is isosceles
hence the angles opposite to equal sides are equal
∠B = ∠C
We also know that sum of angles in a triangle is equal to 180°
∠A + ∠B + ∠C = 180°
2∠B = 180° – 120°
∠B = ∠C = 30°
Вопрос 6. В ΔABC, если AB = AC и ∠B = 70°. Найдите ∠А.
Решение:
In the triangle ABC
Given that ∠B = 70° and AB = AC
According to the question the triangle is isosceles
hence angles opposite to equal sides are equal.
∠B = ∠C
Hence, ∠C = 70°
We also know that sum of angles in a triangle is equal to 180°
∠A + ∠B + ∠C = 180°
∠A = 40°
Hence, ∠A = 40° and ∠C = 70°
Вопрос 7. Вертикальный угол равнобедренного треугольника равен 100°. Найдите его углы при основании.
Решение :
Consider an isosceles ΔABC such that AB = AC
Given that vertical angle, ∠A = 100°
To find the base angles
Since ΔABC is an isosceles triangle hence ∠B = ∠C
We know that sum of interior angles of a triangle = 180°
∠A + ∠B + ∠C = 180°
100° + ∠B +∠B = 180°
2∠B = 180° – 100°
∠B = 40°
∠B = ∠C = 40°
Hence, ∠B = 40° and ∠C = 40°
Вопрос 8. В ∆ABC AB = AC и ∠ACD = 105°. Найдите ∠BAC.
Решение:
Given: AB = AC and ∠ACD = 105°
Since, ∠BCD = 180° = Straight angle
∠BCA + ∠ ACD = 180°
∠BCA + 105° = 180°
Hence, ∠BCA = 75°
Also, ΔABC is an isosceles triangle
AB = AC
∠ABC = ∠ ACB = 75°.
Now the Sum of Interior angles of a triangle = 180°
∠ABC = ∠BCA + ∠CAB = 180°
75° + 75° + ∠CAB = 180°
∠BAC = 30°
Вопрос 9. Найдите величину каждого внешнего угла равностороннего треугольника.
Решение:
We know that for an equilateral triangle
∠ABC = ∠BCA = CAB =180°/3 = 60°
Now,
Extend side BC to D, CA to E, and AB to F.
∠BCA + ∠ACD = 180°
60° + ∠ACD = 180°
∠ACD = 120°
Similarly, we can say that, ∠BAE = ∠FBC = 120°
So, the measure of each exterior angle of an equilateral triangle = 120°.
Вопрос 10. Если основание равнобедренного треугольника образовано с обеих сторон, докажите, что образующиеся при этом внешние углы равны между собой.
Решение:
ED is a straight line segment and B and C are the points on it.
∠EBC = ∠BCD -(Both are equal to 180°)
∠EBA + ∠ABC = ∠ACB + ∠ACD
Since ∠ABC = ∠ACB
Hence, ∠EBA = ∠ACD
Hence, proved
Вопрос 11. По данному рисунку AB = AC и DB = DC найдите отношение ∠ABD:∠ACD.
Решение:
Given: AB = AC, DB = DC
∠ABD = ∠ACD
Hence, ΔABC and ΔDBC are isosceles triangles
∠ABC = ∠ACB and also ∠DBC = ∠DCB (Angles opposite to equal sides are equal)
Now we have to find the ratio =∠ABD : ∠ACD
∠ABD = (∠ABC – ∠DBC)
∠ACD = (∠ACB – ∠DCB)
(∠ABC – ∠DBC):(∠ACB – ∠DCB)
(∠ABC – ∠DBC):(∠ABC – ∠DBC)
1:1
Hence, ∠ ABD:∠ ACD = 1:1
Вопрос 12. Определите величину каждого из равных углов прямоугольного равнобедренного треугольника.
Решение:
ABC is a right-angled triangle
∠A = 90° and AB = AC
Since,
AB = AC
∠C = ∠B
As we know that the sum of angles in a triangle = 180°
∠A + ∠B + ∠C = 180°
90° + ∠ B+ ∠ B = 180°
2∠B = 90°
∠B = 45°
∠B = 45°, ∠C = 45°
So, the measure of each of the equal angles of a right-angled Isosceles triangle = 45°
Вопрос 13. АВ — отрезок. P и Q — точки на противоположных сторонах AB, каждая из которых равноудалена от точек A и B. Покажите, что прямая PQ является серединным перпендикуляром к AB.
Решение:
Given:
AB is a line segment and P, Q are points on opposite sides of AB such that
AP = BP
AQ = BQ
Prove that PQ is the perpendicular bisector of AB.
Let ΔPAQ and ΔPBQ,
AP = BP
AQ = BQ
PQ – PQ
Δ PAQ ≃ Δ PBQ are congruent by SAS congruent condition.
Now, we can observe that APB and ABQ are isosceles triangles.
∠ PAB = ∠ ABQ
And also ∠ QAB = ∠ QBA
Now consider Δ PAC and Δ PBC
C is the point of intersection of AB and PQ
PA = PB
∠APC = ∠BPC
PC = PC
So, by SAS congruency of triangle ΔPAC ≅ ΔPBC.
AC = CB and ∠PCA = ∠PBC
And also, ACB is the line segment
∠ACP + ∠ BCP = 180°
∠ACP = ∠PCB
∠ACP = ∠PCB = 90°
We have AC = CB
C is the midpoint of AB
So, we can conclude that PC is the perpendicular bisector of AB and C is a point on the line PQ.
Therefore, PQ is the perpendicular bisector of AB.