In the previous section we saw the unchanging areas of triangles between parallel lines. We also saw some solved examples. In this section, we will see some more solved examples.
Solved example 14.4
Draw ΔABC with AB = 7 cm, AC = 5 cm, and ∠CAB = 60o. Draw the following three triangles: (i) ΔABP, with ∠BAP = 90o (ii) ΔBCQ, with ∠BCQ = 90o (iii) ΔCAR, with ∠ACR = 90o
Solution:
Two sides and the included angle of ΔABC are given. So we can easily draw it. It is shown in fig.14.7(a) below.
(i) Now we want another ΔABP which satisfies two conditions:
■ It must have the same area as ΔABC
■ It must be right angled at A
• AB is the base. Through the third vertex C, draw a line MN, parallel to AB (shown in fig.b)
• If the new triangle has the same base AB, it's third vertex must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. It must be right angled at A
• For that, draw a line perpendicular to AB at A. This will intersect MN at P
• Join A and B to P. ΔABP is the required triangle
(ii) We want another ΔBCQ which satisfies two conditions:
■ It must have the same area as ΔABC
■ It must be right angled at C
• Take BC as the base. Through the third vertex A, draw a line MN, parallel to BC (shown in fig.c)
• If the new triangle has BC as the base, it's third vertex A must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. It must be right angled at C
• For that, draw a line perpendicular to BC at C. This will intersect MN at Q
• Join B and C to Q. ΔBCQ is the required triangle
(iii) We want another ΔACR which satisfies two conditions:
■ It must have the same area as ΔABC
■ It must be right angled at C
• Take AC as the base. Through the third vertex B, draw a line MN, parallel to AC (shown in fig.d)
• If the new triangle has AC as the base, it's third vertex R must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. It must be right angled at C
• For that, draw a line perpendicular to AC at C. This will intersect MN at R
• Join A and C to R. ΔACR is the required triangle
Solved example 14.5
Draw a circle and a triangle ABC. The triangle should be such that, one of it’s vertex is at the center of the circle, and the other two vertices are on the circumference of the circle. Draw another triangle of the same area as the ABC. All the vertices of the new triangle should be on the circumference of the circle
Solution:
• Fig.14.8(a) shows the circle and the ΔABC. As specified, One of it’s vertex B is at the center of the circle, and the other two vertices A and C are on the circumference.
• We want a new triangle which satisfies the following two conditions:
■ It must have the same area as ABC
■ All vertices of the new triangle must lie on the circumference
• We have vertices A and C already on the circumference. So we can fix AC as the base
• Draw MN parallel to AC, and through the vertex B. This is shown in fig.(b)
• The third vertex of the new triangle should lie on MN. This is to ensure that, the new triangle has the same area as ΔABC
• At the same time, the new vertex should be on the circumference. The point of intersection of MN with the circle will fullfil the two requirements.
• Mark the point of intersection as P. Join A and C to P. ΔACP is the required triangle. This is shown in fig.(c)
Solved example 14.6
Fig.14.9(a) shows a small triangle ABC. Through each vertex of ΔABC, a line is drawn parallel to the corresponding opposite side. Thus the larger triangle PQR is formed. In ΔPQR, 3 additional red lines are drawn as in fig.b. How many triangles are there in fig.b, which have the same area as ΔABC?
Solution:
• We want to find all the triangles which have the same area as ΔABC
• Any triangle which have the same base as AB, BC or CA will have the same area as ΔABC, provided, it’s third vertex lies on the parallel line through the opposite vertex
• Consider the base AC. Any triangle which has the base AC, and having the third vertex on a line parallel to AC, and passing through B, will be having the same area as ABC
• Fig.c shows two such triangles: ΔACQ and ΔACR
• In this way, the other two sides BC and CA have two triangles each. (The reader is advised to find them)
• So, in fig.b, there are a total of six triangles which have the same area as ΔABC
Solved example 14.7
Prove that the two triangles ABC and PQR in fig.14.10(a) are equal in area.
Solution:
• Both the triangles have the same base 4 cm. But heights are not given. So let us mark the heights
• In fig.b, RS ⊥ PQ and CD ⊥ AB. Now we can write the areas:
■ Area of ΔABC = 1⁄2 × b × h = 1⁄2 × 4 × CD
■ Area of ΔPQR = 1⁄2 × b × h = 1⁄2 × 4 × RS - - - (1)
• In fig.b, one more detail is marked: ∠RPS = 50o (∵ ∠RPS and ∠RPQ form a linear pair)
• Now consider the two triangles ΔSPR and ΔADC separately as in fig.c
• ∠SRP = 180 – (90 + 50) = 180 – 140 = 40o (∵ Sum of interior angles of a triangle is 180)
• ∠ACD = 180 – (90 + 50) = 180 – 140 = 40o (∵ Sum of interior angles of a triangle is 180)
■ Thus, in ΔSPR, there are two angles 40o and 50o, with an included side 3 cm
■ In ΔADC also, there are the same two angles 40o and 50o, and with the same included side 3 cm
■ This is a case of ASA congruence
■ Let us write the correspondence:
• 90o is at S and D. So S↔D
• 40o is at R and C. So R↔C
• 50o is at P and A. So P↔A
• So the correspondence is SRP↔DCA
• From this, we can write the corresponding sides: SR↔DC, SP↔DA, RP↔CA
• From the above correspondence of sides, we get SR = DC
• Now look at the equations for areas, that we wrote in (1) above
• When SR = DC, the heights also become equal. So both areas are the same
In the next section we will see the relation between the areas of Quadrilaterals and Triangles.
Solved example 14.4
Draw ΔABC with AB = 7 cm, AC = 5 cm, and ∠CAB = 60o. Draw the following three triangles: (i) ΔABP, with ∠BAP = 90o (ii) ΔBCQ, with ∠BCQ = 90o (iii) ΔCAR, with ∠ACR = 90o
Solution:
Two sides and the included angle of ΔABC are given. So we can easily draw it. It is shown in fig.14.7(a) below.
 |
| Fig.14.7 |
■ It must have the same area as ΔABC
■ It must be right angled at A
• AB is the base. Through the third vertex C, draw a line MN, parallel to AB (shown in fig.b)
• If the new triangle has the same base AB, it's third vertex must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. It must be right angled at A
• For that, draw a line perpendicular to AB at A. This will intersect MN at P
• Join A and B to P. ΔABP is the required triangle
(ii) We want another ΔBCQ which satisfies two conditions:
■ It must have the same area as ΔABC
■ It must be right angled at C
• Take BC as the base. Through the third vertex A, draw a line MN, parallel to BC (shown in fig.c)
• If the new triangle has BC as the base, it's third vertex A must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. It must be right angled at C
• For that, draw a line perpendicular to BC at C. This will intersect MN at Q
• Join B and C to Q. ΔBCQ is the required triangle
(iii) We want another ΔACR which satisfies two conditions:
■ It must have the same area as ΔABC
■ It must be right angled at C
• Take AC as the base. Through the third vertex B, draw a line MN, parallel to AC (shown in fig.d)
• If the new triangle has AC as the base, it's third vertex R must lie on the line MN. This is to ensure that it will have the same area as the original ΔABC
• But having the same area is not enough. It must be right angled at C
• For that, draw a line perpendicular to AC at C. This will intersect MN at R
• Join A and C to R. ΔACR is the required triangle
Solved example 14.5
Draw a circle and a triangle ABC. The triangle should be such that, one of it’s vertex is at the center of the circle, and the other two vertices are on the circumference of the circle. Draw another triangle of the same area as the ABC. All the vertices of the new triangle should be on the circumference of the circle
Solution:
• Fig.14.8(a) shows the circle and the ΔABC. As specified, One of it’s vertex B is at the center of the circle, and the other two vertices A and C are on the circumference.
 |
| Fig.14.8 |
■ It must have the same area as ABC
■ All vertices of the new triangle must lie on the circumference
• We have vertices A and C already on the circumference. So we can fix AC as the base
• Draw MN parallel to AC, and through the vertex B. This is shown in fig.(b)
• The third vertex of the new triangle should lie on MN. This is to ensure that, the new triangle has the same area as ΔABC
• At the same time, the new vertex should be on the circumference. The point of intersection of MN with the circle will fullfil the two requirements.
• Mark the point of intersection as P. Join A and C to P. ΔACP is the required triangle. This is shown in fig.(c)
Solved example 14.6
Fig.14.9(a) shows a small triangle ABC. Through each vertex of ΔABC, a line is drawn parallel to the corresponding opposite side. Thus the larger triangle PQR is formed. In ΔPQR, 3 additional red lines are drawn as in fig.b. How many triangles are there in fig.b, which have the same area as ΔABC?
 |
| Fig.14.9 |
• We want to find all the triangles which have the same area as ΔABC
• Any triangle which have the same base as AB, BC or CA will have the same area as ΔABC, provided, it’s third vertex lies on the parallel line through the opposite vertex
• Consider the base AC. Any triangle which has the base AC, and having the third vertex on a line parallel to AC, and passing through B, will be having the same area as ABC
• Fig.c shows two such triangles: ΔACQ and ΔACR
• In this way, the other two sides BC and CA have two triangles each. (The reader is advised to find them)
• So, in fig.b, there are a total of six triangles which have the same area as ΔABC
Solved example 14.7
Prove that the two triangles ABC and PQR in fig.14.10(a) are equal in area.
 |
| Fig.14.10 |
• Both the triangles have the same base 4 cm. But heights are not given. So let us mark the heights
• In fig.b, RS ⊥ PQ and CD ⊥ AB. Now we can write the areas:
■ Area of ΔABC = 1⁄2 × b × h = 1⁄2 × 4 × CD
■ Area of ΔPQR = 1⁄2 × b × h = 1⁄2 × 4 × RS - - - (1)
• In fig.b, one more detail is marked: ∠RPS = 50o (∵ ∠RPS and ∠RPQ form a linear pair)
• Now consider the two triangles ΔSPR and ΔADC separately as in fig.c
• ∠SRP = 180 – (90 + 50) = 180 – 140 = 40o (∵ Sum of interior angles of a triangle is 180)
• ∠ACD = 180 – (90 + 50) = 180 – 140 = 40o (∵ Sum of interior angles of a triangle is 180)
■ Thus, in ΔSPR, there are two angles 40o and 50o, with an included side 3 cm
■ In ΔADC also, there are the same two angles 40o and 50o, and with the same included side 3 cm
■ This is a case of ASA congruence
■ Let us write the correspondence:
• 90o is at S and D. So S↔D
• 40o is at R and C. So R↔C
• 50o is at P and A. So P↔A
• So the correspondence is SRP↔DCA
• From this, we can write the corresponding sides: SR↔DC, SP↔DA, RP↔CA
• From the above correspondence of sides, we get SR = DC
• Now look at the equations for areas, that we wrote in (1) above
• When SR = DC, the heights also become equal. So both areas are the same
In the next section we will see the relation between the areas of Quadrilaterals and Triangles.
No comments:
Post a Comment