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Practice each question, then open the answer or worked solution.
Show that in a right-angled triangle, the hypotenuse is the longest side.
It is known that ABC is a triangle right angled at B.
We know that,
∠A +∠B+∠C = 180°
Now, if ∠B+∠C = 90° then ∠A has to be 90°.
Since A is the largest angle of the triangle, the side opposite to it must be the largest.
So, AB is the hypotenuse which will be the largest side of the above right-angled triangle i.e. ΔABC.
In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
It is given that ∠PBC < ∠QCB
We know that ∠ABC + ∠PBC = 180°
So, ∠ABC = 180°-∠PBC
Also,
∠ACB +∠QCB = 180°
Therefore ∠ACB = 180° -∠QCB
Now, since ∠PBC < ∠QCB,
∴ ∠ABC > ∠ACB
Hence, AC > AB as sides opposite to the larger angle is always larger.
In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
In the question, it is mentioned that angles B and angle C is smaller than angles A and D respectively i.e. ∠B < ∠A and ∠C < ∠D.
Now,
Since the side opposite to the smaller angle is always smaller
AO < BO — (i)
And OD < OC —(ii)
By adding equation (i) and equation (ii) we get
AO+OD < BO + OC
So, AD < BC
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).
Show that ∠A > ∠C and ∠B > ∠D.
In ΔABD, we see that
AB < AD < BD
So, ∠ADB < ∠ABD — (i) (Since angle opposite to longer side is always larger)
Now, in ΔBCD,
BC < DC < BD
Hence, it can be concluded that
∠BDC < ∠CBD — (ii)
Now, by adding equation (i) and equation (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD
∠ADC < ∠ABC
∠B > ∠D
Similarly, In triangle ABC,
∠ACB < ∠BAC — (iii) (Since the angle opposite to the longer side is always larger)
Now, In ΔADC,
∠DCA < ∠DAC — (iv)
By adding equation (iii) and equation (iv) we get,
∠ACB + ∠DCA < ∠BAC+∠DAC
⇒ ∠BCD < ∠BAD
∴ ∠A > ∠C
In Fig 7.51, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.
It is given that PR > PQ and PS bisects ∠QPR
Now we will have to prove that angle PSR is smaller than PSQ i.e. ∠PSR > ∠PSQ
Proof:
∠QPS = ∠RPS — (ii) (As PS bisects ∠QPR)
∠PQR > ∠PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)
∠PSR = ∠PQR + ∠QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)
By adding (i) and (ii)
∠PQR +∠QPS > ∠PRQ +∠RPS
Thus, from (i), (ii), (iii) and (iv), we get
∠PSR > ∠PSQ
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
First, let “l” be a line segment and “B” be a point lying on it. A line AB perpendicular to l is now drawn. Also, let C be any other point on l. The diagram will be as follows:
To prove:
AB < AC
Proof:
In ΔABC, ∠B = 90°
Now, we know that
∠A+∠B+∠C = 180°
∴ ∠A +∠C = 90°
Hence, ∠C must be an acute angle which implies ∠C < ∠B
So, AB < AC (As the side opposite to the larger angle is always larger)