Using graph, tell whether the pair of linear equations 3x – 5y = 20, 6x – 10y + 40 = 0 is consistent or
inconsistent. Write its solution in case the pair is consistent.
Answers
Answer:
Solution A
If we draw a perpendicular line from the apex angle, it divides the isosceles triangle into two congruent pieces. This can be explained by congruence.
____________________
[Proof]
Consider an isosceles triangle \triangle ABC△ABC with \overline{AB}=\overline{AC}
AB
=
AC
. We draw a perpendicular line \overline{AH}
AH
from the apex angle.
Criterion: SAS
\overline{AB}=\overline{AC}
AB
=
AC
\angle{B}=\angle{C}∠B=∠C
The triangles share \overline{AH}
AH
.
Hence the criterion is satisfied and the two pieces are congruent.
We know that drawing a perpendicular line divides the triangles into two congruent pieces. Since the two pieces are right triangles, we can apply the Pythagorean theorem to find the height.
____________________
[Pythagorean Theorem]
\implies \overline{AH}^2=5^2-\dfrac{5^2}{2^2} =\dfrac{3\cdot 5^2}{2^2}⟹
AH
2
=5
2
−
2
2
5
2
=
2
2
3⋅5
2
\implies \overline{AH}=\boxed{\dfrac{5\sqrt{3} }{2}\ \mathrm{cm}}⟹
AH
=
2
5
3
cm
____________________
[Area of \triangle ABC△ABC ]
Area of a triangle \triangle ABC△ABC with \overline{AB}=5\ \mathrm{cm}, \overline{BC}=5\ \mathrm{cm}, \overline{CA}=5\ \mathrm{cm}
AB
=5 cm,
BC
=5 cm,
CA
=5 cm .
\triangle ABC=\dfrac{1}{2} \times \overline{BC}\times \overline{AH}△ABC=
2
1
×
BC
×
AH
=\dfrac{1}{2} \times 5\times \dfrac{5\sqrt{3} }{2}=\boxed{\dfrac{25\sqrt{3} }{4}\ \mathrm{cm^2}}=
2
1
×5×
2
5
3
=
4
25
3
cm
2
____________________
[Pythagorean Theorem]
\implies \overline{PH}^2=4^2-\dfrac{5^2}{2^2} =\dfrac{64-25}{2^2}=\boxed{\dfrac{39}{2^2} }⟹
PH
2
=4
2
−
2
2
5
2
=
2
2
64−25
=
2
2
39
\implies \overline{PH}=\boxed{\dfrac{\sqrt{39} }{2}\ \mathrm{cm}}⟹
PH
=
2
39
cm
____________________
[Area of \triangle PQR△PQR ]
Area of a triangle \triangle PQR△PQR with \overline{PQ}=4\ \mathrm{cm}, \overline{QR}=5\ \mathrm{cm}, \overline{RP}=4\ \mathrm{cm}
PQ
=4 cm,
QR
=5 cm,
RP
=4 cm .
\triangle PQR=\dfrac{1}{2} \times \overline{QR}\times \overline{PH}△PQR=
2
1
×
QR
×
PH
=\dfrac{1}{2} \times 5\times \dfrac{\sqrt{39} }{2} =\boxed{\dfrac{5\sqrt{39} }{4}\ \mathrm{cm^2}}=
2
1
×5×
2
39
=
4
5
39
cm
2
____________________