Chapter 15.3, Problem 10PSB

(a)

To determine

To Find: The longer base for the given triangle.

Answer to Problem 10PSB

  $\Delta ABC$ has longer base.

Explanation of Solution

Given:

  $\begin{array}{l}\to \Delta ABC\text{ and }\Delta WXY\text{ are isosceles}\\ \to \text{Base}=\overline{AB}\text{ and }\overline{WY}\\ \to \overline{WX}=\overline{BC}\\ \to \angle X=\angle B={50}^o\end{array}$

Formula Used:

The Hinge Theorem: It states that if two sides of two triangles are congruent and angle of one is greater than the angle of second then the third side of the triangle will be greater than the another.

Calculation:

  

  $\begin{array}{l}\text{⇒}m\angle W=m\angle Y=\frac{180-50}{2}=65\\ \text{⇒}m\angle A=m\angle B=50\\ \text{⇒}m\angle C=180-2\left(50\right)=80\\ \text{⇒}m\angle C>m\angle X\\ \text{⇒}AB>WY\left(\overline{WX}\cong \overline{AC}\text{ and }\overline{XY}\cong \overline{BC}\text{, Triangle with longer base has greater opposite angle}\right)\\ \text{⇒}\Delta ABC\text{ has longer base}\end{array}$

Conclusion:

Thus $\Delta ABC$ has longer base

(b)

To determine

To Find: The longer altitude for the given triangle.

Answer to Problem 10PSB

  $\Delta XYW$ has longer altitude

Explanation of Solution

Given:

  $\begin{array}{l}\to \Delta ABC\text{ and }\Delta WXY\text{ are isosceles}\\ \to \text{Base}=\overline{AB}\text{ and }\overline{WY}\\ \to \overline{WX}=\overline{BC}\\ \to \angle X=\angle B={50}^o\end{array}$

Formula Used:

  $\begin{array}{l}\text{The Hinge Theorem : }\\ \text{If two side of one triangle are congruent to two side of the another triangle and the included angle}\\ \text{ in the first triangle is greater than the included angle in the second triangle,then the remaining side}\\ \text{ of the first triangle is greater than the remaining side of the second triangle}\left(\text{SAS}\ne \right).\end{array}$

Calculation:

  

  $\begin{array}{l}\text{⇒}AB>WY\\ \text{⇒}\frac{AB}{2}>\frac{WY}{2}\\ \text{⇒}AD>WV\left(\text{From point a}\right)\\ \text{⇒}W{X}^2=W{V}^2+X{V}^2\left(\text{Pythagoras theorem}\right)\\ \text{⇒}A{C}^2=A{D}^2+C{D}^2\\ \text{⇒}CD<XV\left(WX=AC\text{ and }AD>WV\right)\\ \text{⇒}\Delta XYW\text{ has longer altitude}\end{array}$

Conclusion:

Thus the $\Delta XYW$ has longer altitude